Sets and Counting - Cengage › gpms › websites › 1133228607 › Ch 2... · 2011-09-20 · 70 CHAPTER 2 Sets and Counting Two sets are equalif they contain exactly the same elements.The

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Recently, 1,000 college seniors wereasked whether they favored increasingtheir state’s gasoline tax to generate fundsto improve highways and whether theyfavored increasing their state’s alcohol tax to generate funds to improve thepublic education system. The responseswere tallied, and the following resultswere printed in the campus newspaper:746 students favored an increase in thegasoline tax, 602 favored an increasein the alcohol tax, and 449 favoredincrease in both taxes. How many ofthese 1,000 students favored an increasein at least one of the taxes? How manyfavored increasing only the gasoline tax?How many favored increasing only thealcohol tax? How many favoredincreasing neither tax?

The mathematical tool that wasdesigned to answer questions like these is

WHAT WE WILL DO In This Chapter

WE’LL USE VENN DIAGRAMS TO DEPICT THERELATIONSHIPS BETWEEN SETS:

• One set might be contained within another set.

• Two or more sets might, or might not, shareelements in common.

WE’LL EXPLORE APPLICATIONS OF VENNDIAGRAMS:

• The results of consumer surveys, marketinganalyses, and political polls can be analyzed byusing Venn diagrams.

• Venn diagrams can be used to prove generalformulas related to set theory.

WE’LL EXPLORE VARIOUS METHODS OFCOUNTING:

• A fundamental principle of counting is used todetermine the total number of possible ways ofselecting specified items. For example, how manydifferent student ID numbers are possible at yourschool?

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Sets andCounting

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WHAT WE WILL DO In This Chapter — cont inued

• In selecting items from a specified group, sometimes the order in which theitems are selected matters (the awarding of prizes: first, second, and third),and sometimes it does not (selecting numbers in a lottery or people for acommittee). How does this affect your method of counting?

WE’LL USE SETS IN VARIOUS CONTEXTS:

• In this text, we will use set theory extensively in Chapter 3 on probability.

• Many standardized admissions tests, such as the Graduate Record Exam(GRE) and the Law School Admissions Test (LSAT), ask questions that canbe answered with set theory.

WE’LL EXPLORE SETS THAT HAVE AN INFINITE NUMBER OF ELEMENTS:

• One-to-one correspondences are used to “count” and compare the numberof elements in infinite sets.

• Not all infinite sets have the same number of elements; some infinite setsare countable, and some are not.

the set. Webster’s New World College Dictionary defines a set as “aprescribed collection of points, numbers, or other objects that satisfy agiven condition.” Although you might be able to answer the questionsabout taxes without any formal knowledge of sets, the mental reasoninginvolved in obtaining your answers uses some of the basic principles ofsets. (Incidentally, the answers to the above questions are 899, 297, 153,and 101, respectively.)

The branch of mathematics that deals with sets is called set theory.Set theory can be helpful in solving both mathematical and nonmathematicalproblems. We will explore set theory in the first half of this chapter. Asthe above example shows, set theory often involves the analysis of therelationships between sets and counting the number of elements in aspecific category. Consequently, various methods of counting, collectivelyknown as combinatorics, will be developed and discussed in the secondhalf of this chapter. Finally, what if a set has too many elements to count byusing finite numbers? For example, how many integers are there? Howmany real numbers? The chapter concludes with an exploration of infinitesets and various “levels of infinity.”

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• Not all infinite sets have the same number of elements; some infinite setsLearn

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• Not all infinite sets have the same number of elements; some infinite sets

2.1 Sets and Set Operations

Objectives

• Learn the basic vocabulary and notation of set theory

• Learn and apply the union, intersection, and complement operations

• Draw Venn diagrams

A set is a collection of objects or things. The objects or things in the set are calledelements (or members) of the set. In our example above, we could talk about theset of students who favor increasing only the gasoline tax or the set of students whodo not favor increasing either tax. In geography, we can talk about the set of allstate capitals or the set of all states west of the Mississippi. It is easy to determinewhether something is in these sets; for example, Des Moines is an element of theset of state capitals, whereas Dallas is not. Such sets are called well-definedbecause there is a way of determining for sure whether a particular item is an ele-ment of the set.

EXAMPLE 1 DETERMINING WELL-DEFINED SETS Which of the following sets arewell-defined?

a. the set of all movies directed by Alfred Hitchcockb. the set of all great rock-and-roll bandsc. the set of all possible two-person committees selected from a group of five people

SOLUTION a. This set is well-defined; either a movie was directed by Hitchcock, or it was not.b. This set is not well-defined; membership is a matter of opinion. Some people would say

that the Ramones (one of the pioneer punk bands of the late 1970s) are a member, whileothers might say they are not. (Note: The Ramones were inducted into the Rock andRoll Hall of Fame in 2002.)

c. This set is well-defined; either the two people are from the group of five, or theyare not.

Notation

By tradition, a set is denoted by a capital letter, frequently one that will serve asa reminder of the contents of the set. Roster notation (also called listing nota-tion) is a method of describing a set by listing each element of the set inside thesymbols { and }, which are called set braces. In a listing of the elements of a set,each distinct element is listed only once, and the order of the elements doesn’tmatter.

The symbol � stands for the phrase is an element of, and � stands for isnot an element of. The cardinal number of a set A is the number of elements inthe set and is denoted by n(A). Thus, if R is the set of all letters in the name“Ramones,” then R � {r, a, m, o, n, e, s}. Notice that m is an element of the setR, x is not an element of R, and R has 7 elements. In symbols, m � R, x � R,and n(R) � 7.

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DETERMINING WELL-DEFINED SETS

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whether something is in these sets; for example, Des Moines is an element of theset of state capitals, whereas Dallas is not. Such sets are called

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70 CHAPTER 2 Sets and Counting

Two sets are equal if they contain exactly the same elements. The order inwhich the elements are listed does not matter. If M is the set of all letters in thename “Moaners,” then M � {m, o, a, n, e, r, s}. This set contains exactly thesame elements as the set R of letters in the name “Ramones.” Therefore, M � R �{a, e, m, n, o, r, s}.

Often, it is not appropriate or not possible to describe a set in roster notation.For extremely large sets, such as the set V of all registered voters in Detroit, or forsets that contain an infinite number of elements, such as the set G of all negativereal numbers, the roster method would be either too cumbersome or impossibleto use. Although V could be expressed via the roster method (since each countycompiles a list of all registered voters in its jurisdiction), it would take hundredsor even thousands of pages to list everyone who is registered to vote in Detroit!In the case of the set G of all negative real numbers, no list, no matter how long,is capable of listing all members of the set; there is an infinite number of nega-tive numbers.

In such cases, it is often necessary, or at least more convenient, to useset-builder notation, which lists the rules that determine whether an object isan element of the set rather than the actual elements. A set-builder description ofset G above is

G � {x � x � 0 and x � t}

which is read as “the set of all x such that x is less than zero and x is a real num-ber.” A set-builder description of set V above is

V � {persons � the person is a registered voter in Detroit}

which is read as “the set of all persons such that the person is a registered voter inDetroit.” In set-builder notation, the vertical line stands for the phrase “such that.”

The “Ramones” or The “Moaners”? The set R of all letters in the name “Ramones” is the sameas the set M of all letters in the name “Moaners.” Consequently, the sets are equal; M � R �{a, e, m, n, o, r, s}. (R.I.P. Joey Ramone 1951–2001, Dee Dee Ramone 1952–2002, JohnnyRamone 1948–2004.)

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Whatever is on the left side of the line is the general type of thing in the set, whilethe rules about set membership are listed on the right.

EXAMPLE 2 READING SET-BUILDER NOTATION Describe each of the following inwords.

a. {x � x � 0 and x � t}b. {persons � the person is a living former U.S. president}c. {women � the woman is a former U.S. president}

SOLUTION a. the set of all x such that x is a positive real numberb. the set of all people such that the person is a living former U.S. presidentc. the set of all women such that the woman is a former U.S. president

The set listed in part (c) of Example 2 has no elements; there are no womenwho are former U.S. presidents. If we let W equal “the set of all women such thatthe woman is a former U.S. president,” then n(W ) � 0. A set that has no elementsis called an empty set and is denoted by � or by {}. Notice that since the emptyset has no elements, n(�) � 0. In contrast, the set {0} is not empty; it has one ele-ment, the number zero, so n({0}) � 1.

Universal Set and Subsets

When we work with sets, we must define a universal set. For any given problem, theuniversal set, denoted by U, is the set of all possible elements of any set used in theproblem. For example, when we spell words, U is the set of all letters in the alpha-bet. When every element of one set is also a member of another set, we say that thefirst set is a subset of the second; for instance, {p, i, n} is a subset of {p, i, n, e}.In general, we say that A is a subset of B, denoted by A 8 B, if for every x � A itfollows that x � B. Alternatively, A 8 B if A contains no elements that are not in B.If A contains an element that is not in B, then A is not a subset of B (symbolized asA h B).

EXAMPLE 3 DETERMINING SUBSETS Let B � {countries � the country has a permanentseat on the U.N. Security Council}. Determine whether A is a subset of B.

a. A � {Russian Federation, United States}b. A � {China, Japan}c. A � {United States, France, China, United Kingdom, Russian Federation}d. A � { }

SOLUTION We use the roster method to list the elements of set B.

B � {China, France, Russian Federation, United Kingdom, United States}

a. Since every element of A is also an element of B, A is a subset of B; A 8 B.b. Since A contains an element (Japan) that is not in B, A is not a subset of B; A h B.c. Since every element of A is also an element of B (note that A � B), A is a subset of B

(and B is a subset of A); A 8 B (and B 8 A). In general, every set is a subset of itself;A 8 A for any set A.

d. Does A contain an element that is not in B? No! Therefore, A (an empty set) is a subset ofB; A 8 B. In general, the empty set is a subset of all sets; � 8 A for any set A.

2.1 Sets and Set Operations 71

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We can express the relationship A 8 B visually by drawing a Venn diagram,as shown in Figure 2.1. A Venn diagram consists of a rectangle, representing theuniversal set, and various closed figures within the rectangle, each representing aset. Recall that Venn diagrams were used in Section 1.1 to determine whether anargument was valid.

If two sets are equal, they contain exactly the same elements. It then followsthat each is a subset of the other. For example, if A � B, then every element of A isan element of B (and vice versa). In this case, A is called an improper subset of B.(Likewise, B is an improper subset of A.) Every set is an improper subset of itself;for example, A 8 A. On the other hand, if A is a subset of B and B contains an ele-ment not in A (that is, A � B), then A is called a proper subset of B. To indicate aproper subset, the symbol ( is used. While it is acceptable to write {1, 2} 8 {1, 2, 3},the relationship of a proper subset is stressed when it is written {1, 2} ( {1, 2, 3}.Notice the similarities between the subset symbols, ( and 8, and the inequalitysymbols, � and �, used in algebra; it is acceptable to write 1 � 3, but writing 1 � 3is more informative.

Intersection of Sets

Sometimes an element of one set is also an element of another set; that is, the setsmay overlap. This overlap is called the intersection of the sets. If an element is intwo sets at the same time, it is in the intersection of the sets.


For example, given the sets A � {Buffy, Spike, Willow, Xander} and B �{Angel, Anya, Buffy, Giles, Spike}, their intersection is A ¨ B � {Buffy,Spike}.

Venn diagrams are useful in depicting the relationship between sets. TheVenn diagram in Figure 2.2 illustrates the intersection of two sets; the shadedregion represents A ¨ B.

Mutually Exclusive Sets

Sometimes a pair of sets has no overlap. Consider an ordinary deck of playingcards. Let D � {cards � the card is a diamond} and S � {cards � the card is aspade}. Certainly, no cards are both diamonds and spades at the same time; thatis, S ¨ D � �.

Two sets A and B are mutually exclusive (or disjoint) if they have no ele-ments in common, that is, if A ¨ B � �. The Venn diagram in Figure 2.3 illustratesmutually exclusive sets.

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INTERSECTION OF SETSThe intersection of set A and set B, denoted by A ¨ B, is

A ¨ B � {x � x � A and x � B}

The intersection of two sets consists of those elements that are common toboth sets.

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Union of Sets

What does it mean when we ask, “How many of the 500 college students in a trans-portation survey own an automobile or a motorcycle?” Does it mean “How manystudents own either an automobile or a motorcycle or both?” or does it mean “Howmany students own either an automobile or a motorcycle, but not both?” The for-mer is called the inclusive or, because it includes the possibility of owning both;the latter is called the exclusive or. In logic and in mathematics, the word or refersto the inclusive or, unless you are told otherwise.

The meaning of the word or is important to the concept of union. The unionof two sets is a new set formed by joining those two sets together, just as the unionof the states is the joining together of fifty states to form one nation.

For example, given the sets A � {Conan, David} and B � {Ellen, Katie,Oprah}, their union is A ´ B � {Conan, David, Ellen, Katie, Oprah}, and their in-tersection is A ¨ B � �. Note that because they have no elements in common,A and B are mutually exclusive sets. The Venn diagram in Figure 2.4 illustratesthe union of two sets; the entire shaded region represents A ´ B.

EXAMPLE 4 FINDING THE INTERSECTION AND UNION OF SETS Given the setsA � {1, 2, 3} and B � {2, 4, 6}, find the following.

a. A ¨ B (the intersection of A and B)b. A ´ B (the union of A and B)

SOLUTION a. The intersection of two sets consists of those elements that are common to both sets;therefore, we have

A ¨ B � {1, 2, 3} ¨ {2, 4, 6}

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b. The union of two sets consists of all elements that are in at least one of the sets; there-fore, we have

A ´ B � {1, 2, 3} ´ {2, 4, 6}

� {1, 2, 3, 4, 6}

The Venn diagram in Figure 2.5 shows the composition of each set and illustrates theintersection and union of the two sets.

Because A ´ B consists of all elements that are in A or B (or both), to findn(A ´ B), we add n(A) plus n(B). However, doing so results in an answer thatmight be too big; that is, if A and B have elements in common, these elements willbe counted twice (once as a part of A and once as a part of B). Therefore, to find


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FIGURE 2.5

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The union A ´ B isrepresented by the (entire)shaded region.

FIGURE 2.4

UNION OF SETSThe union of set A and set B, denoted by A ´ B, is

A ´ B � {x � x � A or x � B}

The union of A and B consists of all elements that are in either A or B orboth, that is, all elements that are in at least one of the sets.

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the cardinal number of A ´ B, we add the cardinal number of A to the cardinalnumber of B and then subtract the cardinal number of A ¨ B (so that the overlapis not counted twice).

As long as any three of the four quantities in the general formula are known,the missing quantity can be found by algebraic manipulation.

EXAMPLE 5 ANALYZING THE COMPOSITION OFA UNIVERSAL SET Given n(U) �169, n(A) � 81, and n(B) � 66, find the following.

a. If n(A ¨ B ) � 47, find n(A ´ B ) and draw a Venn diagram depicting the composition ofthe universal set.

b. If n(A ´ B ) � 147, find n(A ¨ B ) and draw a Venn diagram depicting the compositionof the universal set.

SOLUTION a. We must use the Union/Intersection Formula. Substituting the three given quantities, wehave

n(A ´ B) � n(A) � n(B) � n(A ¨ B)

� 81 � 66 � 47

� 100

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n(A ´ B) � n(A) � n(B) � n(A ¨ B)

147 � 81 � 66 � n(A ¨ B)

147 � 147 � n(A ¨ B)

n(A ¨ B) � 147 � 147

n(A ¨ B) � 0

Therefore, A and B have no elements in common; they are mutually exclusive. TheVenn diagram in Figure 2.7 illustrates the composition of U.

CARDINAL NUMBER FORMULA FOR THEUNION/INTERSECTION OF SETSFor any two sets A and B, the number of elements in their union is n(A ´ B),where

n(A ´ B) � n(A) � n(B) � n(A ¨ B)

and n(A ¨ B) is the number of elements in their intersection.

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EXAMPLE 6 ANALYZING THE RESULTS OF A SURVEY A recent transportation surveyof 500 college students (the universal set U) yielded the following information:291 students own an automobile (A), 179 own a motorcycle (M), and 85 own bothan automobile and a motorcycle (A ¨ M). What percent of these students own anautomobile or a motorcycle?

SOLUTION Recall that “automobile or motorcycle” means “automobile or motorcycle or both”(the inclusive or) and that or implies union. Hence, we must find n(A ´ M), thecardinal number of the union of sets A and M. We are given that n(A) � 291,n(M) � 179, and n(A ¨ M) � 85. Substituting the given values into the Union/Intersection Formula, we have

n(A ´ M ) � n(A) � n(M ) � n(A ¨ M )

� 291 � 179 � 85

� 385

Therefore, 385 of the 500 students surveyed own an automobile or a motorcycle.Expressed as a percent, 385�500 � 0.77; therefore, 77% of the students own anautomobile or a motorcycle (or both).

Complement of a Set

In certain situations, it might be important to know how many things are not in agiven set. For instance, when playing cards, you might want to know how manycards are not ranked lower than a five; or when taking a survey, you might want toknow how many people did not vote for a specific proposition. The set of allelements in the universal set that are not in a specific set is called the complementof the set.

For example, given that U � {1, 2, 3, 4, 5, 6, 7, 8, 9} and A � {1, 3, 5, 7, 9}, thecomplement of A is A � {2, 4, 6, 8}. What is the complement of A? Just as�(�x) � x in algebra, (A) � A in set theory. The Venn diagram in Figure 2.8illustrates the complement of set A; the shaded region represents A.

Suppose A is a set of elements, drawn from a universal set U. If x is anelement of the universal set (x � U), then exactly one of the following must betrue: (1) x is an element of A (x � A), or (2) x is not an element of A (x � A).Since no element of the universal set can be in both A and A at the same time,it follows that A and A are mutually exclusive sets whose union equals the en-tire universal set. Therefore, the sum of the cardinal numbers of A and A equalsthe cardinal number of U.


The complement A is repre-sented by the shaded region.

FIGURE 2.8

U

A

COMPLEMENT OF A SETThe complement of set A, denoted by A (read “A prime” or “the complementof A”), is

A � {x � x � U and x � A}

The complement of a set consists of all elements that are in the universal setbut not in the given set.

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John Venn is considered bymany to be one of the

originators of modern symboliclogic. Venn received hisdegree in mathematics fromthe University at Cambridge atthe age of twenty-three. Hewas then elected a fellow ofthe college and held this fellowshipuntil his death, some 66 years later.Two years after receiving his degree,Venn accepted a teaching position atCambridge: college lecturer in moralsciences.

During the latter half of the nineteenthcentury, the study of logic experienced arebirth in England. Mathematicianswere attempting to symbolize and quan-tify the central concepts of logicalthought. Consequently, Venn chose tofocus on the study of logic during histenure at Cambridge. In addition, he in-vestigated the field of probability andpublished The Logic of Chance, his firstmajor work, in 1866.

Venn was well read in the worksof his predecessors, including thenoted logicians Augustus De Morgan,

George Boole, andCharles Dodgson(a.k.a. Lewis Car-roll). Boole’s pioneer-ing work on themarriage of logicand algebra provedto be a strong influ-ence on Venn; infact, Venn used the

type of diagram that nowbears his name in an 1876 paper inwhich he examined Boole’s system ofsymbolic logic.

Venn was not the first scholar to usethe diagrams that now bear his name.Gottfried Leibniz, Leonhard Euler, andothers utilized similar diagrams yearsbefore Venn did. Examining each au-thor’s diagrams, Venn was critical oftheir lack of uniformity. He developed aconsistent, systematic explanation of thegeneral use of geometrical figures in theanalysis of logical arguments. Today,these geometrical figures are known byhis name and are used extensively inelementary set theory and logic.

Venn’s writings were held in highesteem. His textbooks, Symbolic Logic(1881) and The Principles of EmpiricalLogic (1889), were used during the late

nineteenth and early twentieth centuries.In addition to his works on logic andprobability, Venn conducted muchresearch into historical records, espe-cially those of his college and those ofhis family.

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JOHN VENN, 1834–1923

Set theory and the cardinal numbers of setsare used extensively in the study of probability.Although he was a professor of logic, Venninvestigated the foundations and applicationsof theoretical probability. Venn’s first majorwork, The Logic of Chance, exhibited thediversity of his academic interests.

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It is often quicker to count the elements that are not in a set than to countthose that are. Consequently, to find the cardinal number of a set, we can subtractthe cardinal number of its complement from the cardinal number of the universalset; that is, n(A) � n(U) � n(A).

CARDINAL NUMBER FORMULA FOR THECOMPLEMENT OF A SETFor any set A and its complement A,

n(A) � n(A) � n(U)

where U is the universal set.Alternatively,

n(A) � n(U) � n(A) and n(A) � n(U ) � n(A)

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FIGURE 2.11 FIGURE 2.12

FIGURE 2.10

EXAMPLE 7 USING THE COMPLEMENT FORMULA How many letters in the alphabetprecede the letter w?

SOLUTION Rather than counting all the letters that precede w, we will take a shortcut by countingall the letters that do not precede w. Let L � {letters � the letter precedes w}. Therefore, L � {letter � the letter does not precede w}. Now L � {w, x, y, z}, andn(L) � 4; therefore, we have

n(L) � n(U) � n(L) Complement Formula� 26 � 4� 22

There are twenty-two letters preceding the letter w.

Shading Venn Diagrams

In an effort to visualize the results of operations on sets, it may be necessary toshade specific regions of a Venn diagram. The following example shows a system-atic method for shading the intersection or union of any two sets.

EXAMPLE 8 SHADING VENN DIAGRAMS On a Venn diagram, shade in the region cor-responding to the indicated set.

a. A ¨ B b. A ´ B

SOLUTION a. First, draw and label two overlapping circles as shown in Figure 2.9. The two “compo-nents” of the operation A ¨ B are “A” and “B.” Shade each of these components incontrasting ways; shade one of them, say A, with horizontal lines, and the other withvertical lines as in Figure 2.10. Be sure to include a legend, or key, identifying each typeof shading.

To be in the intersection of two sets, an element must be in both sets at the sametime. Therefore, the intersection of A and B is the region that is shaded in bothdirections (horizontal and vertical) at the same time. A final diagram depicting A ¨ B isshown in Figure 2.11.

U

A B

Two overlapping circles.

FIGURE 2.9

b. Refer to Figure 2.10. To be in the union of two sets, an element must be in at least oneof the sets. Therefore, the union of A and B consists of all regions that are shaded in anydirection whatsoever (horizontal or vertical or both). A final diagram depicting A ´ Bis shown in Figure 2.12.

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SHADING VENN DIAGRAMS

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Set Theory LogicCommon

Term Symbol Term Symbol Wording

union ´ disjunction ∨ or

intersection ¨ conjunction ∧ and

complement negation � not

subset 8 conditional S if . . . then . . .

Comparison of terms and symbols used in set theory and logic.FIGURE 2.13

1. State whether the given set is well defined.

a. the set of all black automobiles

b. the set of all inexpensive automobiles

c. the set of all prime numbers

d. the set of all large numbers

2. Suppose A � {2, 5, 7, 9, 13, 25, 26}.

a. Find n(A)

b. True or false: 7 � A

c. True or false: 9 � A

d. True or false: 20 � A

2.1 Exercises

Basic Operationsin Set Theory Logical Biconditional

union [x � (A ´ B)] 4 [x � A ∨ x � B]

intersection [x � (A ¨ B)] 4 [x � A ∧ x � B]

complement (x � A) 4 � (x � A)

subset (A 8 B) 4 (x � A S x � B)

Set theory operations as logical biconditionals.FIGURE 2.14

Set Theory and Logic

If you have read Chapter 1, you have probably noticed that set theory and logichave many similarities. For instance, the union symbol ´ and the disjunction sym-bol ∨ have the same meaning, but they are used in different circumstances; ´ goesbetween sets, while ∨ goes between logical expressions. The ´ and ∨ symbols aresimilar in appearance because their usages are similar. A comparison of the termsand symbols used in set theory and logic is given in Figure 2.13.

Applying the concepts and symbols of Chapter 1, we can define the basic opera-tions of set theory in terms of logical biconditionals. The biconditionals in Figure 2.14are tautologies (expressions that are always true); the first biconditional is read as “x is an element of the union of sets A and B if and only if x is an element of set A or x is an element of set B.”

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2.1 Exercises 79

27. Suppose n(U) � 150, n(A) � 37, and n(B) � 84.

a. If n(A ´ B) � 100, find n(A ¨ B) and draw a Venndiagram illustrating the composition of U.

b. If n(A ´ B) � 121, find n(A ¨ B) and draw a Venndiagram illustrating the composition of U.

28. Suppose n(U) � w, n(A) � x, n(B) � y, and

n(A ´ B) � z.

a. Why must x be less than or equal to z?

b. If A � U and B � U, fill in the blank with the mostappropriate symbol: �, �, �, or .

w_____z, w_____y, y_____z, x_____w

c. Find n(A ¨ B) and draw a Venn diagram illustratingthe composition of U.

29. In a recent transportation survey, 500 high school sen-iors were asked to check the appropriate box or boxeson the following form:

I own an automobile.I own a motorcycle.

The results were tabulated as follows: 102 studentschecked the automobile box, 147 checked the motor-cycle box, and 21 checked both boxes.

a. Draw a Venn diagram illustrating the results of thesurvey.

b. What percent of these students own an automobileor a motorcycle?

30. In a recent market research survey, 500 marriedcouples were asked to check the appropriate boxor boxes on the following form:

We own a DVD player.We own a microwave oven.

The results were tabulated as follows: 301 coupleschecked the DVD player box, 394 checked themicrowave oven box, and 217 checked both boxes.


b. What percent of these couples own a DVD player ora microwave oven?

31. In a recent socioeconomic survey, 700 married womenwere asked to check the appropriate box or boxes onthe following form:

I have a career.I have a child.

In Exercises 3–6, list all subsets of the given set. Identify whichsubsets are proper and which are improper.

3. B � {Lennon, McCartney}4. N � {0}5. S � {yes, no, undecided}6. M � {classical, country, jazz, rock}

In Exercises 7–10, the universal set is U � {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.

7. If A � {1, 2, 3, 4, 5} and B � {4, 5, 6, 7, 8}, find thefollowing.

a. A ¨ B b. A ´ B

c. A d. B

8. If A � {2, 3, 5, 7} and B � {2, 4, 6, 7}, find thefollowing.

a. A ¨ B b. A ´ B

c. A d. B

9. If A � {1, 3, 5, 7, 9} and B � {0, 2, 4, 6, 8}, find thefollowing.

a. A ¨ B b. A ´ B

c. A d. B

10. If A � {3, 6, 9} and B � {4, 8}, find the following.

a. A ¨ B b. A ´ B

c. A d. B

In Exercises 11–16, the universal set is U � {Monday, Tuesday,Wednesday, Thursday, Friday, Saturday, Sunday}. IfA � {Monday, Tuesday, Wednesday, Thursday, Friday} andB � {Friday, Saturday, Sunday}, find the indicated set.

11. A ¨ B 12. A ´ B

13. B 14. A

15. A ´ B 16. A ¨ B

In Exercises 17–26, use a Venn diagram like the one in Figure 2.15 to shade in the region corresponding to theindicated set.

17. A ¨ B 18. A ´ B

19. A 20. B

21. A ´ B 22. A ´ B

23. A ¨ B 24. A ¨ B

25. A ´ B 26. A ¨ B

U

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44. face cards or diamonds

45. face cards and black

46. face cards and diamonds

47. aces or 8’s 48. 3’s or 6’s

49. aces and 8’s 50. 3’s and 6’s

51. Suppose A � {1, 2, 3} and B � {1, 2, 3, 4, 5, 6}.

a. Find A ¨ B.

b. Find A ´ B.

c. In general, if E ¨ F � E, what must be trueconcerning sets E and F?

d. In general, if E ´ F � F, what must be trueconcerning sets E and F?

52. Fill in the blank, and give an example to support youranswer.

a. If A ( B, then A ¨ B � _____.

b. If A ( B, then A ´ B � _____.

53. a. List all subsets of A � {a}. How many subsets doesA have?

b. List all subsets of A � {a, b}. How many subsetsdoes A have?

c. List all subsets of A � {a, b, c}. How many subsetsdoes A have?

d. List all subsets of A � {a, b, c, d}. How manysubsets does A have?

e. Is there a relationship between the cardinal number ofset A and the number of subsets of set A?

f. How many subsets does A � {a, b, c, d, e, f} have?

HINT: Use your answer to part (e).

54. Prove the Cardinal Number Formula for the Comple-ment of a Set.

HINT:Apply the Union/Intersection Formula to A and A.

Answer the following questions using completesentences and your own words.

• Concept Questions

55. If A ¨ B � �, what is the relationship between setsA and B?

56. If A ´ B � �, what is the relationship between setsA and B?

57. Explain the difference between {0} and �.

58. Explain the difference between 0 and {0}.

59. Is it possible to have A ¨ A � �?

60. What is the difference between proper and impropersubsets?

61. Aset can be described by two methods: the roster methodand set-builder notation. When is it advantageous to usethe roster method? When is it advantageous to use set-builder notation?

The results were tabulated as follows: 285 womenchecked the child box, 316 checked the career box, and196 were blank (no boxes were checked).


b. What percent of these women had both a child anda career?

32. In a recent health survey, 700 single men in theirtwenties were asked to check the appropriate box orboxes on the following form:

I am a member of a private gym. I am a vegetarian.

The results were tabulated as follows: 349 menchecked the gym box, 101 checked the vegetarian box,and 312 were blank (no boxes were checked).


b. What percent of these men were both members of aprivate gym and vegetarians?

For Exercises 33–36, let

U � {x � x is the name of one of the states in the United States}

A � {x � x � U and x begins with the letter A}

I � {x � x � U and x begins with the letter I}

M � {x � x � U and x begins with the letter M}

N � {x � x � U and x begins with the letter N}

O � {x � x � U and x begins with the letter O}

33. Find n(M). 34. Find n(A ´ N ).

35. Find n(I ¨ O). 36. Find n(M ¨ I ).

For Exercises 37–40, let

U � {x � x is the name of one of the months in a year}

J � {x � x � U and x begins with the letter J}

Y � {x � x � U and x ends with the letter Y}

V � {x � x � U and x begins with a vowel}

R � {x � x � U and x ends with the letter R}

37. Find n(R). 38. Find n(J ¨ V ).

39. Find n(J ´ Y ). 40. Find n(V ¨ R).

In Exercises 41–50, determine how many cards, in an ordinarydeck of fifty-two, fit the description. (If you are unfamiliar withplaying cards, see the end of Section 3.1 for a description of astandard deck.)

41. spades or aces 42. clubs or 2’s

43. face cards or black

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2.2 Applications of Venn Diagrams 81

65. Which of the following pairs of groups selected byJohn and Juneko conform to the restrictions?

John Juneko

a. J, K, L M, N, O

b. J, K, P L, M, N

c. K, N, P J, M, O

d. L, M, N K, O, P

e. M, O, P J, K, N

66. If N is in John’s group, which of the following couldnot be in Juneko’s group?

a. J b. K c. L d. M e. P

67. If K and N are in John’s group, Juneko’s group mustconsist of which of the following?

a. J, M, and O

b. J, O, and P

c. L, M, and P

d. L, O, and P

e. M, O, and P

68. If J is in Juneko’s group, which of the following is true?

a. K cannot be in John’s group.

b. N cannot be in John’s group.

c. O cannot be in Juneko’s group.

d. P must be in John’s group.

e. P must be in Juneko’s group.

69. If K is in John’s group, which of the following is true?

a. J must be in John’s group.

b. O must be in John’s group.

c. L must be in Juneko’s group.

d. N cannot be in John’s group.

e. O cannot be in Juneko’s group.

62. Translate the following symbolic expressions intoEnglish sentences.

a. x � (A ¨ B) 4 (x � A ∧ x � B)

b. (x � A) 4 � (x � A)

c. (A 8 B) 4 (x � A S x � B)

• history Questions

63. In what academic field was John Venn a professor?Where did he teach?

64. What was one of John Venn’s main contributions to thefield of logic? What new benefits did it offer?

Exercises 65–69 refer to the following: Two collectors, John andJuneko, are each selecting a group of three posters from a groupof seven movie posters: J, K, L, M, N, O, and P. No poster can bein both groups. The selections made by John and Juneko aresubject to the following restrictions:

● If K is in John’s group, M must be in Juneko’s group.● If N is in John’s group, P must be in Juneko’s group.● J and P cannot be in the same group.● M and O cannot be in the same group.

THE NEXT LEVELIf a person wants to pursue an advanced degree(something beyond a bachelor’s or four-yeardegree), chances are the person must take a stan-dardized exam to gain admission to a school or tobe admitted into a specific program. These examsare intended to measure verbal, Quantitative, andanalytical skills that have developed throughout aperson’s life. Many classes and study guides areavailable to help people prepare for the exams.The following questions are typical of thosefound in the study guides.

2.2 Applications of Venn Diagrams

Objectives

• Use Venn diagrams to analyze the results of surveys

• Develop and apply De Morgan’s Laws of complements

As we have seen, Venn diagrams are very useful tools for visualizing the relation-ships between sets. They can be used to establish general formulas involving setoperations and to determine the cardinal numbers of sets. Venn diagrams are par-ticularly useful in survey analysis.

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Surveys

Surveys are often used to divide people or objects into categories. Because the cat-egories sometimes overlap, people can fall into more than one category. Venn dia-grams and the formulas for cardinal numbers can help researchers organize the data.

EXAMPLE 1 ANALYZING THE RESULTS OF A SURVEY: TWO SETS Has the adventof the DVD affected attendance at movie theaters? To study this question,Professor Redrum’s film class conducted a survey of people’s movie-watchinghabits. He had his students ask hundreds of people between the ages of sixteen andforty-five to check the appropriate box or boxes on the following form:

After the professor had collected the forms and tabulated the results, he told theclass that 388 people had checked the theater box, 495 had checked the DVD box,281 had checked both boxes, and 98 of the forms were blank. Giving the class onlythis information, Professor Redrum posed the following three questions.

a. What percent of the people surveyed watched a movie in a theater or on a DVD duringthe past month?

b. What percent of the people surveyed watched a movie in a theater only?c. What percent of the people surveyed watched a movie on a DVD only?

SOLUTION a. To calculate the desired percentages, we must determine n(U), the total number of peo-ple surveyed. This can be accomplished by drawing a Venn diagram. Because the sur-vey divides people into two categories (those who watched a movie in a theater andthose who watched a movie on a DVD), we need to define two sets. Let

T � {people � the person watched a movie in a theater}

D � {people � the person watched a movie on a DVD}

Now translate the given survey information into the symbols for the sets and attach theirgiven cardinal numbers: n(T) � 388, n(D) � 495, and n(T ¨ D) � 281.

Our first goal is to find n(U ). To do so, we will fill in the cardinal numbers of allregions of a Venn diagram consisting of two overlapping circles (because we aredealing with two sets). The intersection of T and D consists of 281 people, so we drawtwo overlapping circles and fill in 281 as the number of elements in common (seeFigure 2.16).

Because we were given n(T) � 388 and know that n(T ¨ D) � 281, the difference388 � 281 � 107 tells us that 107 people watched a movie in a theater but did notwatch a movie on a DVD. We fill in 107 as the number of people who watched a movieonly in a theater (see Figure 2.17).

Because n(D) � 495, the difference 495 � 281 � 214 tells us that 214 peoplewatched a movie on a DVD but not in a theater. We fill in 214 as the number of peoplewho watched a movie only on a DVD (see Figure 2.18).

The only region remaining to be filled in is the region outside both circles. Thisregion represents people who didn’t watch a movie in a theater or on a DVD andis symbolized by (T ´ D). Because 98 people didn’t check either box on the form,n[(T ´ D)] � 98 (see Figure 2.19).

After we have filled in the Venn diagram with all the cardinal numbers, we readily seethat n(U) � 98 � 107 � 281 � 214 � 700. Therefore, 700 people were in the survey.

I watched a movie in a theater during the past month.I watched a movie on a DVD during the past month.

U

T D281

n(T ¨ D) � 281.

FIGURE 2.16

U

T

107281

D

n(T ´ D) � 107.

FIGURE 2.17

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I watched a movie in a theater during the past month.

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To determine what percent of the people surveyed watched a movie in a theater oron a DVD during the past month, simply divide n(T ´ D) by n(U):

� 0.86

Therefore, exactly 86% of the people surveyed watched a movie in a theater or ona DVD during the past month.

b. To find what percent of the people surveyed watched a movie in a theater only, divide107 (the number of people who watched a movie in a theater only) by n(U):

L 0.153 (rounding off to three decimal places)

Approximately 15.3% of the people surveyed watched a movie in a theater only.c. Because 214 people watched a movie on DVD only, 214�700 � 0.305714285 . . . , or

approximately 30.6%, of the people surveyed watched a movie on DVD only.

When you solve a cardinal number problem (a problem that asks, “Howmany?” or “What percent?”) involving a universal set that is divided into variouscategories (for instance, a survey), use the following general steps.

107

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U

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FIGURE 2.19

U

T

107281

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D

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FIGURE 2.18

U

A B

C

Three overlapping circles.

FIGURE 2.20When we are working with three sets, we must account for all possible inter-

sections of the sets. Hence, in such cases, we will use the Venn diagram shown inFigure 2.20

EXAMPLE 2 ANALYZING THE RESULTS OF A SURVEY: THREE SETS A consumersurvey was conducted to examine patterns in ownership of notebook computers,cellular telephones, and DVD players. The following data were obtained:213 people had notebook computers, 294 had cell phones, 337 had DVD players,109 had all three, 64 had none, 198 had cell phones and DVD players, 382 had cellphones or notebook computers, and 61 had notebook computers and DVD playersbut no cell phones.

a. What percent of the people surveyed owned a notebook computer but no DVD player orcell phone?

SOLVING A CARDINAL NUMBER PROBLEMA cardinal number problem is a problem in which you are asked, “Howmany?” or “What percent?”1. Define a set for each category in the universal set. If a category and its negation

are both mentioned, define one set A and utilize its complement A.2. Draw a Venn diagram with as many overlapping circles as the number of sets

you have defined.3. Write down all the given cardinal numbers corresponding to the various given sets.4. Starting with the innermost overlap, fill in each region of the Venn diagram

with its cardinal number.5. In answering a “what percent” problem, round off your answer to the nearest

tenth of a percent.

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b. What percent of the people surveyed owned a DVD player but no notebook computer orcell phone?

SOLUTION a. To calculate the desired percentages, we must determine n(U), the total number ofpeople surveyed. This can be accomplished by drawing a Venn diagram. Because thesurvey divides people into three categories (those who own a notebook computer,those who own a cell phone, and those who own a DVD player), we need to definethree sets. Let

C � {people � the person owns a notebook computer}

T � {people � the person owns a cellular telephone}

D � {people � the person owns a DVD player}

Now translate the given survey information into the symbols for the sets and attach theirgiven cardinal numbers:

Our first goal is to find n(U). To do so, we will fill in the cardinal numbers of allregions of a Venn diagram like that in Figure 2.20. We start by using informationconcerning membership in all three sets. Because the intersection of all three setsconsists of 109 people, we fill in 109 in the region common to C and T and D (seeFigure 2.21).

Next, we utilize any information concerning membership in two of the three sets. Be-cause n(T ¨ D) � 198, a total of 198 people are common to both T and D; some are in C,and some are not in C. Of these 198 people, 109 are in C (see Figure 2.21). Therefore, thedifference 198 � 109 � 89 gives the number not in C. Eighty-nine people are in T and Dand not in C; that is, n(T ¨ D ¨ C) � 89. Concerning membership in the two sets C andD, we are given n(C ¨ D ¨ T) � 61. Therefore, we know that 61 people are in C and Dand not in T (see Figure 2.22).

We are given n(T ´ C) � 382. From this number, we can calculate n(T ¨ C) byusing the Union/Intersection Formula:

n(T ´ C) � n(T ) � n(C) – n(T ¨ C)

382 � 294 � 213 – n(T ¨ C)

n(T ¨ C) � 125

Therefore, a total of 125 people are in T and C; some are in D, and some are notin D. Of these 125 people, 109 are in D (see Figure 2.21). Therefore, the difference

U

109

C T

D

U

10961 89

C T

D

n(C ¨ T ¨ D) � 109.

FIGURE 2.21

Determining cardinal numbersin a Venn diagram.

FIGURE 2.22

213 people had notebook computers ¬¬¬¬¬¬¬S n(C) � 213

294 had cellular telephones ¬¬¬¬¬¬¬¬¬¬¬S n(T) � 294

337 had DVD players ¬¬¬¬¬¬¬¬¬¬¬¬¬S n(D) � 337

109 had all three ¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬S n(C ¨ T ¨ D) � 109(C and T and D)

64 had none ¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬S n(C� ¨ T� ¨ D� ) � 64 (not C and not T and not D)

198 had cell phones and DVD players ¬¬¬¬¬¬S n(T ¨ D) � 198 (T and D)

382 had cell phones or notebook computers ¬¬¬¬S n(T ´ C) � 382 (T or C)

61 had notebook computers and DVD playersbut no cell phones ¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬S n(C ¨ D ¨ T) � 61 (C and D and not T)

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FIGURE 2.23 Determining cardinal numbersin a Venn diagram.

FIGURE 2.24

U

10980

16

61 89

C T

D

125 � 109 � 16 gives the number not in D. Sixteen people are in T and C and notin D; that is, n(C ¨ T ¨ D) � 16 (see Figure 2.23).

Knowing that a total of 294 people are in T (given n(T) � 294), we are now able tofill in the last region of T. The missing region (people in T only) has 294 � 109 � 89 �16 � 80 members; n(T ¨ C ¨ D) � 80 (see Figure 2.24).

In a similar manner, we subtract the known pieces of C from n(C) � 213, which isgiven, and obtain 213 � 61 � 109 � 16 � 27; therefore, 27 people are in C only.Likewise, to find the last region of D, we use n(D) � 337 (given) and obtain 337 � 89 �109 � 61 � 78; therefore, 78 people are in D only. Finally, the 64 people who own noneof the items are placed “outside” the three circles (see Figure 2.25).

By adding up the cardinal numbers of all the regions in Figure 2.25, we find that thetotal number of people in the survey is 524; that is, n(U) � 524.

Now, to determine what percent of the people surveyed owned only a notebookcomputer (no DVD player and no cell phone), we simply divide n(C ¨ D ¨ T) byn(U):

� 0.051526717 . . .

Approximately 5.2% of the people surveyed owned a notebook computer and did notown a DVD player or a cellular telephone.

b. To determine what percent of the people surveyed owned only a DVD player (no note-book computer and no cell phone), we divide n(D ¨ C ¨ T) by n(U):

� 0.148854961. . .

Approximately 14.9% of the people surveyed owned a DVD player and did not own anotebook computer or a cell phone.

De Morgan’s Laws

One of the basic properties of algebra is the distributive property:

Given a(b � c), the operation outside the parentheses can be distributed over theoperation inside the parentheses. It makes no difference whether you add b and cfirst and then multiply the sum by a or first multiply each pair, a and b, a and c, andthen add their products; the same result is obtained. Is there a similar property forthe complement, union, and intersection of sets?

a1b � c 2 � ab � ac

n1D º Cº T 2n1U 2 �

78

524

n1C º Dº T 2n1U 2 �

27

524

A completed Venn diagram.

FIGURE 2.25

U

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61

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27

89

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EXAMPLE 3 INVESTIGATING THE COMPLEMENT OF A UNION Suppose U �{1, 2, 3, 4, 5}, A � {1, 2, 3}, and B � {2, 3, 4}.

a. For the given sets, does (A ´ B) � A ´ B?b. For the given sets, does (A ´ B) � A ¨ B?

SOLUTION a. To find (A ´ B), we must first find A ´ B:

The complement of A ´ B (relative to the given universal set U) is

To find A ´ B, we must first find A and B:

The union of A and B is

Now, {5} � {1, 4, 5}; therefore, (A ´ B) � A ´ B.b. We find (A ´ B) as in part (a): (A ´ B) � {5}. Now,

For the given sets, (A ´ B) � A ¨ B.

Part (a) of Example 3 shows that the operation of complementation cannot beexplicitly distributed over the operation of union; that is, (A ´ B) � A ´ B.However, part (b) of the example implies that there may be some relationshipbetween the complement, union, and intersection of sets. The fact that (A ´ B) �A ¨ B for the given sets A and B does not mean that it is true for all sets A and B.We will use a general Venn diagram to examine the validity of the statement (A ´ B) � A ¨ B.

When we draw two overlapping circles within a universal set, four regionsare formed. Every element of the universal set U is in exactly one of the followingregions, as shown in Figure 2.26:

I in neither A nor BII in A and not in BIII in both A and BIV in B and not in A

The set A ´ B consists of all elements in regions II, III, and IV. Therefore, thecomplement (A ´ B) consists of all elements in region I. A consists of all ele-ments in regions I and IV, and B consists of the elements in regions I and II. There-fore, the elements common to both A and B are those in region I; that is, the setA ¨ B consists of all elements in region I. Since (A ´ B) and A ¨ B contain ex-actly the same elements (those in region I), the sets are equal; that is, (A ´ B) �A ¨ B is true for all sets A and B.

The relationship (A ´ B) � A ¨ B is known as one of De Morgan’sLaws. Simply stated, “the complement of a union is the intersection of the com-plements.” In a similar manner, it can be shown that (A ¨ B) � A ´ B (seeExercise 33).

� 556A � B � 54, 56 � 51, 56

� 51, 4, 56A ´ B � 54, 56 ´ 51, 56

A � 54, 56 and B � 51, 56

1A ´ B 2 � 556

� 51, 2, 3, 46A ´ B � 51, 2, 36 ´ 52, 3, 46

Four regions in a universal set U.

FIGURE 2.26

U

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95057_02_ch02_p067-130.qxd 9/27/10 9:52 AM Page 86

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Being born blind in one eyedid not stop Augustus De

Morgan from becoming awell-read philosopher, histo-rian, logician, and mathemati-cian. De Morgan was born inMadras, India, where his father wasworking for the East India Company.On moving to England, De Morganwas educated at Cambridge, and at theage of twenty-two, he became the firstprofessor of mathematics at the newlyopened University of London (later re-named University College).

De Morgan viewed all of mathemat-ics as an abstract study of symbols andof systems of operations applied to thesesymbols. While studying the ramifica-tions of symbolic logic, De Morgan for-mulated the general properties ofcomplementation that now bear hisname. Not limited to symbolic logic,De Morgan’s many works include booksand papers on the foundations of alge-bra, differential calculus, and probabil-ity. He was known to be a jovial personwho was fond of puzzles, and his wittyand amusing book A Budget of Para-doxes still entertains readers today.Besides his accomplishments in the aca-demic arena, De Morgan was an ex-pert flutist, spoke five languages, andthoroughly enjoyed big-city life.

Knowing of his interest in probabil-ity, an actuary (someone who studieslife expectancies and determinespayments of premiums for insurance

companies) onceasked De Morgan aquestion concerningthe probability that acertain group of peo-ple would be alive ata certain time. In his re-sponse, De Morganemployed a formula

containing the number p. In amaze-ment, the actuary responded, “Thatmust surely be a delusion! What can acircle have to do with the number ofpeople alive at a certain time?” DeMorgan replied that p has numerousapplications and occurrences in manydiverse areas of mathematics. Becauseit was first defined and used in geome-try, people are conditioned to acceptthe mysterious number only in referenceto a circle. However, in the history ofmathematics, if probability had beensystematically studied before geometryand circles, our present-day inter-pretation of the number p would beentirely different. In addition to his ac-complishments in logic and higher-levelmathematics, De Morgan introduced aconvention with which we are all familiar:In a paper written in 1845, he suggestedthe use of a slanted line to representa fraction, such as 1�2 or 3�4.

De Morgan was a staunch defenderof academic freedom and religious tol-erance. While he was a student atCambridge, his application for a fellow-ship was refused because he would nottake and sign a theological oath. Laterin life, he resigned his professorship as aprotest against religious bias. (University

College gave preferential treatment tomembers of the Church of Englandwhen textbooks were selected and didnot have an open policy on religiousphilosophy.) Augustus De Morgan wasa man who was unafraid to take a standand make personal sacrifices when itcame to principles he believed in.

HistoricalNote

AUGUSTUS DE MORGAN, 1806–1871

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Gematria is a mystic pseudoscience in whichnumbers are substituted for the letters in a name. DeMorgan’s book A Budget of Paradoxes containsseveral gematria puzzles, such as, “Mr. Davis Thomfound a young gentleman of the name of St. Clairebusy at the Beast number: he forthwith added theletters in �� (the Greek spelling of St. Claire)and found 666.” (Verify this by using the Greeknumeral system.)

DE MORGAN’S LAWSFor any sets A and B,

(A ´ B) � A ¨ B

That is, the complement of a union is the intersection of the complements. Also,

(A ¨ B) � A ´ B

That is, the complement of an intersection is the union of the complements.

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Human blood types are a classicexample of set theory. As you may

know, there are four categories (or sets)of blood types: A, B, AB, and O. Know-ing someone’s blood type is extremelyimportant in case a blood transfusion isrequired; if blood of two different typesis combined, the blood cells may beginto clump together, with potentially fatalconsequences! (Do you know yourblood type?)

What exactly are “blood types”? Inthe early 1900s, the Austrian scientistKarl Landsteiner observed the presence(or absence) of two distinct chemicalmolecules on the surface of all red bloodcells in numerous samples of humanblood. Consequently, he labeled onemolecule “A” and the other “B.” Thepresence or absence of these specificmolecules is the basis of the universalclassification of blood types. Specifi-cally, blood samples containing only theA molecule are labeled type A, whereasthose containing only the B molecule arelabeled type B. If a blood sample con-tains both molecules (A and B) it is la-beled type AB; and if neither is present,the blood is typed as O. The presence(or absence) of these molecules can bedepicted in a standard Venn diagram asshown in Figure 2.27. In the notation ofset operations, type A blood is denotedA ¨ B, type B is B ¨ A, type AB is A ¨ B, and type O is A ¨ B.

If a specific blood sample is mixedwith blood containing a blood molecule(A or B) that it does not already have,

Topic x

the presence of the foreign moleculemay cause the mixture of blood toclump. For example, type A blood can-not be mixed with any blood containingthe B molecule (type B or type AB).Therefore, a person with type A bloodcan receive a transfusion only of type Aor type O blood. Consequently, a per-son with type AB blood may receive atransfusion of any blood type; type AB isreferred to as the “universal receiver.”Because type O blood contains neitherthe A nor the B molecule, all blood typesare compatible with type O blood; typeO is referred to as the “universal donor.”

It is not uncommon for scientists tostudy rhesus monkeys in an effort to learnmore about human physiology. In sodoing, a certain blood protein was dis-covered in rhesus monkeys. Subse-quently, scientists found that the blood ofsome people contained this protein,whereas the blood of others did not. Thepresence, or absence, of this protein in

human blood is referred to as the Rhfactor; blood containing the protein is la-beled “Rh+”, whereas “Rh–” indicatesthe absence of the protein. The Rh factorof human blood is especially importantfor expectant mothers; a fetus can de-velop problems if its parents have oppo-site Rh factors.

When a person’s blood is typed, thedesignation includes both the regularblood type and the Rh factor. For in-stance, type AB– indicates the presenceof both the A and B molecules (type AB),along with the absence of the rhesusprotein; type O+ indicates the absenceof both the A and B molecules (type O),along with the presence of the rhesusprotein. Utilizing the Rh factor, there areeight possible blood types as shown inFigure 2.28.

We will investigate the occurrenceand compatibility of the various bloodtypes in Example 5 and in Exercises35–43.

Blood types and thepresence of the Aand B molecules.

FIGURE 2.27 Blood types combinedwith the Rh factor.

FIGURE 2.28

BLOOD TYPES: SET THEORY IN THE REAL WORLD

O

ABA B

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EXAMPLE 4 APPLYING DE MORGAN’S LAW Suppose U � {0, 1, 2, 3, 4, 5, 6, 7, 8, 9},A � {2, 3, 7, 8}, and B � {0, 4, 5, 7, 8, 9}. Use De Morgan’s Law to find (A ´ B).

SOLUTION The complement of a union is equal to the intersection of the complements; there-fore, we have

(A ´ B) � (A) ¨ B De Morgan’s Law� A ¨ B (A�)� � A� {2, 3, 7, 8} ¨ {1, 2, 3, 6}� {2, 3}

95057_02_ch02_p067-130.qxd 9/27/10 9:52 AM Page 88

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Notice that this problem could be done without using De Morgan’s Law, butsolving it would then involve finding first A, then A ´ B, and finally (A ´ B).This method would involve more work. (Try it!)

EXAMPLE 5 INVESTIGATING BLOOD TYPES IN THE UNITED STATES TheAmerican Red Cross has compiled a massive database of the occurrence ofblood types in the United States. Their data indicate that on average, out ofevery 100 people in the United States, 44 have the A molecule, 15 have the Bmolecule, and 45 have neither the A nor the B molecule. What percent of the U.S. population have the following blood types?

a. Type O? b. Type AB? c. Type A? d. Type B?

SOLUTION a. First, we define the appropriate sets. Let

A � {Americans � the person has the A molecule}

B � {Americans � the person has the B molecule}

We are given the following cardinal numbers: n(U) � 100, n(A) � 44, n(B) � 15, andn(A � B) � 45. Referring to Figure 2.27, and given that 45 people (out of 100) haveneither the A molecule nor the B molecule, we conclude that 45 of 100 people, or 45%,have type O blood as shown in Figure 2.29.

b. Applying De Morgan’s Law to the Complement Formula, we have the following.

n(A � B) � n[(A � B)] � n(U) Complement Formulan(A � B) � n(A � B) � n(U) applying De Morgan’s Law

n(A � B) � 45 � 100 substituting known values

Therefore, n(A � B) � 55.Now, use the Union/Intersection Formula.

n(A � B) � n(A) � n(B) � n(A � B) Union/Intersection Formula

55 � 44 � 15 � n(A � B) substituting known values

n(A � B) � 44 � 15 � 55 adding n(A � B) and subtracting 55

Therefore, n(A � B) � 4. This means that 4 people (of 100) have both the Aand the B molecules; that is, 4 of 100 people, or 4%, have type AB blood. SeeFigure 2.30.

c. Knowing that a total of 44 people have the A molecule, that is, n(A) � 44, we subtractn(A � B) � 4 and conclude that 40 have only the A molecule.

Therefore, n(A � B) � 40. This means that 40 people (of 100) have only theA molecule; that is, 40 of 100 people, or 40%, have type A blood. See Figure 2.31.

d. Knowing that a total of 15 people have the B molecule, that is, n(B) � 15, we subtractn(A � B) � 4 and conclude that 11 have only the B molecule.

Therefore, n(B � A) � 11. This means that 11 people (of 100) have only theB molecule; that is, 11 of 100 people, or 11%, have type B blood (see Figure 2.32).

45

O

ABA B

Forty-five of 100 (45%) havetype O blood.

FIGURE 2.29

45

O

AB

4A B

Four in 100 (4%) have typeAB blood.

FIGURE 2.30

45

O

AB

4B

40

A

45

40

O

AB

4 11

A B

Forty in 100 (40%)have type A blood.

FIGURE 2.31 Eleven in 100 (11%)have type B blood.

FIGURE 2.32

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Applying De Morgan’s Law to the Complement Formula, we have the following.

The occurrence of blood types in the United States is summarized in Figure 2.33.


Blood Type O A B AB

Occurrence 45% 40% 11% 4%

Occurrence of blood types in the United States.FIGURE 2.33

1. A survey of 200 people yielded the followinginformation: 94 people owned a DVD player, 127owned a microwave oven, and 78 owned both. Howmany people owned the following?

a. a DVD player or a microwave oven

b. a DVD player but not a microwave oven

c. a microwave oven but not a DVD player

d. neither a DVD player nor a microwave oven

2. A survey of 300 workers yielded the followinginformation: 231 workers belonged to a union, and 195were Democrats. If 172 of the union members wereDemocrats, how many workers were in the followingsituations?

a. belonged to a union or were Democrats

b. belonged to a union but were not Democrats

c. were Democrats but did not belong to a union

d. neither belonged to a union nor were Democrats

3. The records of 1,492 high school graduates were exam-ined, and the following information was obtained:1,072 graduates took biology, and 679 took geometry.If 271 of those who took geometry did not takebiology, how many graduates took the following?

a. both classes

b. at least one of the classes

c. biology but not geometry

d. neither class

4. A department store surveyed 428 shoppers, and thefollowing information was obtained: 214 shoppersmade a purchase, and 299 were satisfied with theservice they received. If 52 of those who made apurchase were not satisfied with the service, how manyshoppers did the following?

a. made a purchase and were satisfied with the service

b. made a purchase or were satisfied with the service

c. were satisfied with the service but did not make apurchase

d. were not satisfied and did not make a purchase

5. In a survey, 674 adults were asked what televisionprograms they had recently watched. The followinginformation was obtained: 226 adults watched neitherthe Big Game nor the New Movie, and 289 watchedthe New Movie. If 183 of those who watched the NewMovie did not watch the Big Game, how many of thesurveyed adults watched the following?

a. both programs

b. at least one program

c. the Big Game

d. the Big Game but not the New Movie

2.2 Exercises

Blood types per 100 people in the United States. (Source: American Red Cross.)FIGURE 2.34

Blood Type O+ O– A+ A– B+ B– AB+ AB–

Occurrence 38 7 34 6 9 2 3 1

The occurrence of blood types given in Figure 2.33 can be further catego-rized by including the Rh factor. According to the American Red Cross, out ofevery 100 people in the United States, blood types and Rh factors occur at the ratesshown in Figure 2.34.

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telephones, and DVD players. The following datawere obtained: 313 people had laptop computers,232 had cell phones, 269 had DVD players, 69 had allthree, 64 had none, 98 had cell phones and DVDplayers, 57 had cell phones but no computers or DVDplayers, and 104 had computers and DVD players butno cell phones.

a. What percent of the people surveyed owned a cellphone?

b. What percent of the people surveyed owned onlya cell phone?

12. In a recent survey of monetary donations made bycollege graduates, the following information wasobtained: 95 graduates had donated to a politicalcampaign, 76 had donated to assist medical research,133 had donated to help preserve the environment, 25had donated to all three, 22 had donated to none of thethree, 38 had donated to a political campaign andto medical research, 46 had donated to medicalresearch and to preserve the environment, and 54 haddonated to a political campaign and to preserve theenvironment.

a. What percent of the college graduates donated tonone of the three listed causes?

b. What percent of the college graduates donated toexactly one of the three listed causes?

13. Recently, Green Day, the Kings of Leon, and the BlackEyed Peas had concert tours in the United States. Alarge group of college students was surveyed, and thefollowing information was obtained: 381 students sawBlack Eyed Peas, 624 saw the Kings of Leon, 712 saw Green Day, 111 saw all three, 513 saw none, 240 sawonly Green Day, 377 saw Green Day and the Kings ofLeon, and 117 saw the Kings of Leon and Black EyedPeas but not Green Day.

a. What percent of the college students saw at leastone of the bands?

b. What percent of the college students saw exactlyone of the bands?

14. Dr. Hawk works in an allergy clinic, and his patientshave the following allergies: 68 patients are allergic todairy products, 93 are allergic to pollen, 91 are allergicto animal dander, 31 are allergic to all three, 29 areallergic only to pollen, 12 are allergic only to dairyproducts, and 40 are allergic to dairy products andpollen.

a. What percent of Dr. Hawk’s patients are allergic toanimal dander?

b. What percent of Dr. Hawk’s patients are allergiconly to animal dander?

15. When the members of the Eye and I Photo Clubdiscussed what type of film they had used duringthe past month, the following information wasobtained: 77 members used black and white, 24 usedonly black and white, 65 used color, 18 used only

6. A survey asked 816 college freshmen whether theyhad been to a movie or eaten in a restaurant during thepast week. The following information was obtained:387 freshmen had been to neither a movie nor arestaurant, and 266 had been to a movie. If 92 of thosewho had been to a movie had not been to a restaurant,how many of the surveyed freshmen had been to thefollowing?

a. both a movie and a restaurant

b. a movie or a restaurant

c. a restaurant

d. a restaurant but not a movie

7. A local 4-H club surveyed its members, and thefollowing information was obtained: 13 members hadrabbits, 10 had goats, 4 had both rabbits and goats, and18 had neither rabbits nor goats.

a. What percent of the club members had rabbits orgoats?

b. What percent of the club members had only rabbits?

c. What percent of the club members had only goats?

8. A local anime fan club surveyed its members regardingtheir viewing habits last weekend, and the followinginformation was obtained: 30 members had watched anepisode of Naruto, 44 had watched an episode ofDeath Note, 21 had watched both an episode of Narutoand an episode of Death Note, and 14 had watchedneither Naruto nor Death Note.

a. What percent of the club members had watchedNaruto or Death Note?

b. What percent of the club members had watchedonly Naruto?

c. What percent of the club members had watchedonly Death Note?

9. A recent survey of w shoppers (that is, n(U) � w)yielded the following information: x shoppersshopped at Sears, y shopped at JCPenney’s, and zshopped at both. How many people shopped at thefollowing?

a. Sears or JCPenney’s

b. only Sears

c. only JCPenney’s

d. neither Sears nor JCPenney’s

10. A recent transportation survey of w urban commuters(that is, n(U) � w) yielded the following information:x commuters rode neither trains nor buses, y rodetrains, and z rode only trains. How many people rodethe following?

a. trains and buses

b. only buses

c. buses

d. trains or buses

11. A consumer survey was conducted to examinepatterns in ownership of laptop computers, cellular

2.2 Exercises 91

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20. In a recent health insurance survey, employeesat a large corporation were asked, “Have you beena patient in a hospital during the past year, and if so,for what reason?” The following results wereobtained: 494 employees had an injury, 774 had anillness, 1,254 had tests, 238 had an injury and an ill-ness and tests, 700 had an illness and tests, 501 hadtests and no injury or illness, 956 had an injury orillness, and 1,543 had not been a patient.

a. What percent of the employees had been patients ina hospital?

b. What percent of the employees had tests in ahospital?

In Exercises 21 and 22, use a Venn diagram like the one inFigure 2.35.

color, 101 used black and white or color, 27 usedinfrared, 9 used all three types, and 8 didn’t use anyfilm during the past month.

a. What percent of the members used only infraredfilm?

b. What percent of the members used at least two ofthe types of film?

16. After leaving the polls, many people are asked howthey voted. (This is called an exit poll.) ConcerningPropositions A, B, and C, the following informationwas obtained: 294 people voted yes on A, 90 votedyes only on A, 346 voted yes on B, 166 voted yesonly on B, 517 voted yes on A or B, 339 voted yes onC, no one voted yes on all three, and 72 voted no onall three.

a. What percent of the voters in the exit poll voted noon A?

b. What percent of the voters voted yes on more thanone proposition?

17. In a recent survey, consumers were asked where theydid their gift shopping. The following results wereobtained: 621 consumers shopped at Macy’s, 513shopped at Emporium, 367 shopped at Nordstrom,723 shopped at Emporium or Nordstrom, 749shopped at Macy’s or Nordstrom, 776 shopped atMacy’s or Emporium, 157 shopped at all three,96 shopped at neither Macy’s nor Emporium norNordstrom.

a. What percent of the consumers shopped at morethan one store?

b. What percent of the consumers shopped exclusivelyat Nordstrom?

18. A company that specializes in language tutoring liststhe following information concerning its English-speaking employees: 23 employees speak German; 25speak French; 31 speak Spanish; 43 speak Spanish orFrench; 38 speak French or German; 46 speak Germanor Spanish; 8 speak Spanish, French, and German; and7 speak English only.

a. What percent of the employees speak at least onelanguage other than English?

b. What percent of the employees speak at least twolanguages other than English?

19. In a recent survey, people were asked which radiostation they listened to on a regular basis. The followingresults were obtained: 140 people listened to WOLD(oldies), 95 listened to WJZZ (jazz), 134 listened toWTLK (talk show news), 235 listened to WOLD orWJZZ, 48 listened to WOLD and WTLK, 208 listenedto WTLK or WJZZ, and 25 listened to none.

a. What percent of people in the survey listened onlyto WTLK on a regular basis?

b. What percent of people in the survey did not listento WTLK on a regular basis?


21. A survey of 136 pet owners yielded the followinginformation: 49 pet owners own fish; 55 own a bird;50 own a cat; 68 own a dog; 2 own all four; 11 ownonly fish; 14 own only a bird; 10 own fish and a bird;21 own fish and a cat; 26 own a bird and a dog; 27 owna cat and a dog; 3 own fish, a bird, a cat, and no dog;1 owns fish, a bird, a dog, and no cat; 9 own fish, a cat,a dog, and no bird; and 10 own a bird, a cat, a dog, andno fish. How many of the surveyed pet owners have nofish, no birds, no cats, and no dogs? (They own othertypes of pets.)

22. An exit poll of 300 voters yielded the followinginformation regarding voting patterns on PropositionsA, B, C, and D: 119 voters voted yes on A; 163 votedyes on B; 129 voted yes on C; 142 voted yes on D; 37voted yes on all four; 15 voted yes on A only; 50 votedyes on B only; 59 voted yes on A and B; 70 voted yeson A and C; 82 voted yes on B and D; 93 voted yes onC and D; 10 voted yes on A, B, and C and no on D; 2voted yes on A, B, and D and no on C; 16 voted yes onA, C, and D and no on B; and 30 voted yes on B, C, andD and no on A. How many of the surveyed voters votedno on all four propositions?

U

A

B C

D

Four overlapping regions.FIGURE 2.35

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40. If a person has type A blood, what blood types may theperson receive in a transfusion?

41. If a person has type B blood, what blood types may theperson receive in a transfusion?

42. If a person has type AB blood, what blood types maythe person receive in a transfusion?

43. If a person has type O blood, what blood types may theperson receive in a transfusion?


• History Questions

44. What notation did De Morgan introduce in regard tofractions?

45. Why did De Morgan resign his professorship atUniversity College?

Exercises 46–52 refer to the following: A nonprofit organization’sboard of directors, composed of four women (Angela, Betty,Carmen, and Delores) and three men (Ed, Frank, and Grant),holds frequent meetings. A meeting can be held at Betty’s house, atDelores’s house, or at Frank’s house.

● Delores cannot attend any meetings at Betty’s house.

● Carmen cannot attend any meetings on Tuesday or onFriday.

● Angela cannot attend any meetings at Delores’s house.

● Ed can attend only those meetings that Grant also attends.

● Frank can attend only those meetings that both Angela andCarmen attend.

46. If all members of the board are to attend a particularmeeting, under which of the following circumstancescan it be held?

a. Monday at Betty’s

b. Tuesday at Frank’s

c. Wednesday at Delores’s

d. Thursday at Frank’s

e. Friday at Betty’s

THE NEXT LEVELIf a person wants to pursue an advanced degree(something beyond a bachelor’s or four-yeardegree), chances are the person must take a stan-dardized exam to gain admission to a school or tobe admitted into a specific program. These examsare intended to measure verbal, quantitative, andanalytical skills that have developed throughout aperson’s life. Many classes and study guides areavailable to help people prepare for the exams.The following questions are typical of thosefound in the study guides.

In Exercises 23–26, given the sets U � {0, 1, 2, 3, 4, 5, 6, 7, 8, 9},A � {0, 2, 4, 5, 9}, and B � {1, 2, 7, 8, 9}, use De Morgan’s Lawsto find the indicated sets.

23. (A ´ B)

24. (A ¨ B)

25. (A ¨ B)

26. (A ´ B)

In Exercises 27–32, use a Venn diagram like the one in Figure 2.36to shade in the region corresponding to the indicated set.

2.2 Exercises 93

27. A ¨ B ¨ C

28. A ´ B ´ C

29. (A ´ B) ¨ C

30. A ¨ (B ´ C)

31. B ¨ (A ´ C)

32. (A ´ B) ¨ C

33. Using Venn diagrams, prove De Morgan’s Law (A ¨ B) � A ´ B.

34. Using Venn diagrams, prove A ´ (B ¨ C) �(A ´ B) ¨ (A ´ C).

Use the data in Figure 2.34 to complete Exercises 35–39. Roundoff your answers to a tenth of a percent.

35. What percent of all people in the United States haveblood that is

a. Rh positive? b. Rh negative?

36. Of all people in the United States who have type Oblood, what percent are


37. Of all people in the United States who have type Ablood, what percent are


38. Of all people in the United States who have type Bblood, what percent are


39. Of all people in the United States who have type ABblood, what percent are


U

A B

C

Three overlapping circles.FIGURE 2.36

95057_02_ch02_p067-130.qxd 9/27/10 9:52 AM Page 93

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History Questions

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What notation did De Morgan introduce in regard to

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What notation did De Morgan introduce in regard to

Why did De Morgan resign his professorship at

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Why did De Morgan resign his professorship atUniversity College?

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University College?

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THE NEXT LEVELLearn

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THE NEXT LEVELLearn

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51. If Grant is unable to attend a meeting on Tuesday atDelores’s, what is the largest possible number of boardmembers who can attend?

a. 1 b. 2 c. 3

d. 4 e. 5

52. If a meeting is held on Friday, which of the followingboard members cannot attend?

a. Grant b. Delores c. Ed

d. Betty e. Frank

Web Project

53. A person’s Rh factor will limit the person’s optionsregarding the blood types he or she may receive duringa transfusion. Fill in the following chart. How does aperson’s Rh factor limit that person’s options regardingcompatible blood?

Some useful links for this web project are listed on thetext web site:www.cengage.com/math/johnson

47. Which of the following can be the group that attends ameeting on Wednesday at Betty’s?

a. Angela, Betty, Carmen, Ed, and Frank

b. Angela, Betty, Ed, Frank, and Grant

c. Angela, Betty, Carmen, Delores, and Ed

d. Angela, Betty, Delores, Frank, and Grant

e. Angela, Betty, Carmen, Frank, and Grant

48. If Carmen and Angela attend a meeting but Grant isunable to attend, which of the following could betrue?

a. The meeting is held on Tuesday.

b. The meeting is held on Friday.

c. The meeting is held at Delores’s.

d. The meeting is held at Frank’s.

e. The meeting is attended by six of the boardmembers.

49. If the meeting is held on Tuesday at Betty’s, which ofthe following pairs can be among the board memberswho attend?

a. Angela and Frank

b. Ed and Betty

c. Carmen and Ed

d. Frank and Delores

e. Carmen and Angela

50. If Frank attends a meeting on Thursday that is not heldat his house, which of the following must be true?

a. The group can include, at most, two women.

b. The meeting is at Betty’s house.

c. Ed is not at the meeting.

d. Grant is not at the meeting.

e. Delores is at the meeting.


If Your Blood Type Is: You Can Receive:

O+

O–

A+

A–

B+

B–

AB+

AB–

2.3 Introduction to Combinatorics

Objectives

• Develop and apply the Fundamental Principle of Counting

• Develop and evaluate factorials

If you went on a shopping spree and bought two pairs of jeans, three shirts, and twopairs of shoes, how many new outfits (consisting of a new pair of jeans, a new shirt,and a new pair of shoes) would you have? A compact disc buyers’ club sends youa brochure saying that you can pick any five CDs from a group of 50 of today’s

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regarding the blood types he or she may receive duringa transfusion. Fill in the following chart. How does a

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a transfusion. Fill in the following chart. How does aperson’s Rh factor limit that person’s options regarding

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person’s Rh factor limit that person’s options regarding

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hottest sounds for only $1.99. How many different combinations can you choose?Six local bands have volunteered to perform at a benefit concert, and there is someconcern over the order in which the bands will perform. How many different line-ups are possible? The answers to questions like these can be obtained by listing allthe possibilities or by using three shortcut counting methods: the FundamentalPrinciple of Counting, combinations, and permutations. Collectively, thesemethods are known as combinatorics. (Incidentally, the answers to the questionsabove are 12 outfits, 2,118,760 CD combinations, and 720 lineups.) In this section,we consider the first shortcut method.

The Fundamental Principle of Counting

Daily life requires that we make many decisions. For example, we must decidewhat food items to order from a menu, what items of clothing to put on inthe morning, and what options to order when purchasing a new car. Often, weare asked to make a series of decisions: “Do you want soup or salad? Whattype of dressing? What type of vegetable? What entrée? What beverage?What dessert?” These individual components of a complete meal lead tothe question “Given all the choices of soups, salads, dressings, vegetables,entrées, beverages, and desserts, what is the total number of possible dinnercombinations?”

When making a series of decisions, how can you determine the total num-ber of possible selections? One way is to list all the choices for each category andthen match them up in all possible ways. To ensure that the choices are matchedup in all possible ways, you can construct a tree diagram. A tree diagram con-sists of clusters of line segments, or branches, constructed as follows: A clusterof branches is drawn for each decision to be made such that the number ofbranches in each cluster equals the number of choices for the decision. For in-stance, if you must make two decisions and there are two choices for decision 1and three choices for decision 2, the tree diagram would be similar to the oneshown in Figure 2.37.

2.3 Introduction to Combinatorics 95

choice #1

choice #1

decision 2

decision 1

decision 2

decision 2

choice #1

choice #2

choice #2

choice #2

choice #3

choice #3

A tree diagram.FIGURE 2.37

Although this method can be applied to all problems, it is very time consum-ing and impractical when you are dealing with a series of many decisions, each ofwhich contains numerous choices. Instead of actually listing all possibilities via atree diagram, using a shortcut method might be desirable. The following examplegives a clue to finding such a shortcut.

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the morning, and what options to order when purchasing a new car. Often, weare asked to make a series of decisions: “Do you want soup or salad? What

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are asked to make a series of decisions: “Do you want soup or salad? Whattype of dressing? What type of vegetable? What entrée? What beverage?

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entrées, beverages, and desserts, what is the total number of possible dinner

EXAMPLE 1 DETERMINING THE TOTAL NUMBER OF POSSIBLE CHOICES IN ASERIES OF DECISIONS If you buy two pairs of jeans, three shirts, and twopairs of shoes, how many new outfits (consisting of a new pair of jeans, a new shirt,and a new pair of shoes) would you have?

SOLUTION Because there are three categories, selecting an outfit requires a series of threedecisions: You must select one pair of jeans, one shirt, and one pair of shoes. We willmake our three decisions in the following order: jeans, shirt, and shoes. (The orderin which the decisions are made does not affect the overall outfit.)

Our first decision (jeans) has two choices ( jeans 1 or jeans 2); our tree startswith two branches, as in Figure 2.38.

Our second decision is to select a shirt, for which there are three choices. Ateach pair of jeans on the tree, we draw a cluster of three branches, one for eachshirt, as in Figure 2.39.

Our third decision is to select a pair of shoes, for which there are two choices.At each shirt on the tree, we draw a cluster of two branches, one for each pair ofshoes, as in Figure 2.40.


start

shirt 1shoes 1

Possible Outfits

jeans 1, shirt 1, shoes 1












shoes 2

shoes 1

shoes 2

shoes 1

shoes 2

shoes 1

shoes 2

shoes 1

shoes 2

shoes 1

shoes 2

shirt 2

shirt 3

shirt 1

shirt 2

shirt 3

jeans 1

jeans 2

The third decision.FIGURE 2.40

We have now listed all possible ways of putting together a new outfit; twelveoutfits can be formed from two pairs of jeans, three shirts, and two pairs of shoes.

Referring to Example 1, note that each time a decision had to be made, thenumber of branches on the tree diagram was multiplied by a factor equal to thenumber of choices for the decision. Therefore, the total number of outfits couldhave been obtained by multiplying the number of choices for each decision:

jeans 2

jeans 1

start

The first decision.

FIGURE 2.38

shirt 1

shirt 2

shirt 3

jeans 2

start

shirt 1

shirt 2

shirt 3

jeans 1

The second decision.

FIGURE 2.39

2 · 3 · 2 � 12jeans ⎯⎯↑ ↑ ↑ ↑⎯ outfitsshirts ⎯⎯⎯⏐ ⏐shoes ⎯⎯⎯⎯⎯⏐

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The generalization of this process of multiplication is called the Fundamental Prin-ciple of Counting.


EXAMPLE 2 APPLYING THE FUNDAMENTAL PRINCIPLE OF COUNTING A serialnumber consists of two consonants followed by three nonzero digits followed by avowel (A, E, I, O, U): for example, “ST423E” and “DD666E.” Determine howmany serial numbers are possible given the following conditions.

a. Letters and digits cannot be repeated in the same serial number.b. Letters and digits can be repeated in the same serial number.

SOLUTION a. Because the serial number has six symbols, we must make six decisions. Consequently,we must draw six boxes:

There are twenty-one different choices for the first consonant. Because the letters can-not be repeated, there are only twenty choices for the second consonant. Similarly, thereare nine different choices for the first nonzero digit, eight choices for the second, andseven choices for the third. There are five different vowels, so the total number of pos-sible serial numbers is

� � � � � � 1,058,400

consonants nonzero digits vowel

There are 1,058,400 possible serial numbers when the letters and digits cannot be re-peated within a serial number.

b. Because letters and digits can be repeated, the number of choices does not decrease byone each time as in part (a). Therefore, the total number of possibilities is

� � � � � � 1,607,445

consonants nonzero digits vowel

There are 1,607,445 possible serial numbers when the letters and digits can be repeatedwithin a serial number.

Factorials

EXAMPLE 3 APPLYING THE FUNDAMENTAL PRINCIPLE OF COUNTING Threestudents rent a three-bedroom house near campus. One of the bedrooms is verydesirable (it has its own bath), one has a balcony, and one is undesirable (it is verysmall). In how many ways can the housemates choose the bedrooms?

↑↑ ↑↑ ↑59992121

↑↑ ↑↑ ↑57892021

THE FUNDAMENTAL PRINCIPLE OF COUNTINGThe total number of possible outcomes of a series of decisions (makingselections from various categories) is found by multiplying the number ofchoices for each decision (or category) as follows:1. Draw a box for each decision.2. Enter the number of choices for each decision in the appropriate box and

multiply.

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SOLUTION Three decisions must be made: who gets the room with the bath, who gets the roomwith the balcony, and who gets the small room. Using the Fundamental Principleof Counting, we draw three boxes and enter the number of choices for each decision.There are three choices for who gets the room with the bath. Once that decision hasbeen made, there are two choices for who gets the room with the balcony, and finally,there is only one choice for the small room.

� � � 6

There are six different ways in which the three housemates can choose the threebedrooms.

Combinatorics often involve products of the type 3 · 2 · 1 � 6, as seen inExample 3. This type of product is called a factorial, and the product 3 · 2 · 1 iswritten as 3!. In this manner, 4! � 4 · 3 · 2 · 1 (� 24), and 5! � 5 · 4 · 3 · 2 · 1 (� 120).

123


Many scientific calculators have a button that will calculate a factorial.Depending on your calculator, the button will look like or , and you mighthave to press a or button first. For example, to calculate 6!, type thenumber 6, press the factorial button, and obtain 720. To calculate a factorial onmost graphing calculators, do the following:

• Type the value of n. (For example, type the number 6.)• Press the button.• Press the right arrow button as many times as necessary to highlight .• Press the down arrow as many times as necessary to highlight the “!” symbol, and

press .• Press to execute the calculation.

To calculate a factorial on a Casio graphing calculator, do the following:

• Press the button; this gives you access to the main menu.• Press 1 to select the RUN mode; this mode is used to perform arithmetic operations.• Type the value of n. (For example, type the number 6.)• Press the button; this gives you access to various options displayed at the

bottom of the screen.• Press the button to see more options (i.e., ).• Press the button to select probability options (i.e., ).• Press the button to select factorial (i.e., ).• Press the button to execute the calculation.

The factorial symbol “n!” was first introduced by Christian Kramp (1760–1826) ofStrasbourg in his Élements d’Arithmétique Universelle (1808). Before theintroduction of this “modern” symbol, factorials were commonly denoted by mn.However, printing presses of the day had difficulty printing this symbol; conse-quently, the symbol n! came into prominence because it was relatively easy fora typesetter to use.

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FACTORIALSIf n is a positive integer, then n factorial, denoted by n!, is the product of allpositive integers less than or equal to n.

n! � n · (n – 1) · (n – 2) · · · · · 2 · 1

As a special case, we define 0! � 1.

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EXAMPLE 4 EVALUATING FACTORIALS Find the following values.

a. 6! b. c.

SOLUTION a. 6! � 6 · 5 · 4 · 3 · 2 · 1� 720

Therefore, 6! � 720.

8!

3! · 5!

8!

5!

c.

Therefore, � 56.

Using a calculator, we obtain the same result.

8!3! · 5!

� 56

�8 · 7 · 6 3 · 2 · 1

�8 · 7 · 6 · 5 · 4 · 3 · 2 · 113 · 2 · 1 2 15 · 4 · 3 · 2 · 1 2

8!

3! · 5!�

8 · 7 · 6 · 5 · 4 · 3 · 2 · 113 · 2 · 1 2 15 · 4 · 3 · 2 · 1 2

b.

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8!5! � 336.

� 336 � 8 · 7 · 6

�8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 5 · 4 · 3 · 2 · 1

8!

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8 · 7 · 6 · 5 · 4 · 3 · 2 · 15 · 4 · 3 · 2 · 1

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1. A nickel, a dime, and a quarter are tossed.

a. Use the Fundamental Principle of Counting todetermine how many different outcomes arepossible.

b. Construct a tree diagram to list all possible out-comes.

2. A die is rolled, and a coin is tossed.

a. Use the Fundamental Principle of Counting todetermine how many different outcomes arepossible.

b. Construct a tree diagram to list all possibleoutcomes.

3. Jamie has decided to buy either a Mega or a BetterByte desktop computer. She also wants to purchaseeither Big Word, Word World, or Great Word word-processing software and either Big Number or NumberWorld spreadsheet software.

a. Use the Fundamental Principle of Counting todetermine how many different packages of acomputer and software Jamie has to choose from.

b. Construct a tree diagram to list all possiblepackages of a computer and software.

4. Sammy’s Sandwich Shop offers a soup, sandwich, andbeverage combination at a special price. There are threesandwiches (turkey, tuna, and tofu), two soups(minestrone and split pea), and three beverages (coffee,milk, and mineral water) to choose from.

a. Use the Fundamental Principle of Counting todetermine how many different meal combinationsare possible.

b. Construct a tree diagram to list all possible soup,sandwich, and beverage combinations.

5. If you buy three pairs of jeans, four sweaters, and twopairs of boots, how many new outfits (consisting of anew pair of jeans, a new sweater, and a new pair ofboots) will you have?

6. A certain model of automobile is available in sixexterior colors, three interior colors, and three interiorstyles. In addition, the transmission can be eithermanual or automatic, and the engine can have eitherfour or six cylinders. How many different versions ofthe automobile can be ordered?

7. To fulfill certain requirements for a degree, a studentmust take one course each from the following groups:health, civics, critical thinking, and elective. If thereare four health, three civics, six critical thinking, andten elective courses, how many different options forfulfilling the requirements does a student have?

8. To fulfill a requirement for a literature class, a studentmust read one short story by each of the following

authors: Stephen King, Clive Barker, Edgar Allan Poe,and H. P. Lovecraft. If there are twelve King, sixBarker, eight Poe, and eight Lovecraft stories to choosefrom, how many different combinations of readingassignments can a student choose from to fulfill thereading requirement?

9. A sporting goods store has fourteen lines of snow skis,seven types of bindings, nine types of boots, and threetypes of poles. Assuming that all items are compatiblewith each other, how many different complete skiequipment packages are available?

10. An audio equipment store has ten different amplifiers,four tuners, six turntables, eight tape decks, six compactdisc players, and thirteen speakers. Assuming that allcomponents are compatible with each other, how manydifferent complete stereo systems are available?

11. A cafeteria offers a complete dinner that includes oneserving each of appetizer, soup, entrée, and dessert for$6.99. If the menu has three appetizers, four soups, sixentrées, and three desserts, how many different mealsare possible?

12. A sandwich shop offers a “U-Chooz” special consistingof your choice of bread, meat, cheese, and special sauce(one each). If there are six different breads, eight meats,five cheeses, and four special sauces, how manydifferent sandwiches are possible?

13. How many different Social Security numbers arepossible? (A Social Security number consists of ninedigits that can be repeated.)

14. To use an automated teller machine (ATM), a customermust enter his or her four-digit Personal IdentificationNumber (PIN). How many different PINs are possible?

15. Every book published has an International StandardBook Number (ISBN). The number is a code used toidentify the specific book and is of the form X-XXX-XXXXX-X, where X is one of digits 0, 1, 2, . . . , 9. How many different ISBNs are possible?

16. How many different Zip Codes are possible using(a) the old style (five digits) and (b) the new style (ninedigits)? Why do you think the U.S. Postal Serviceintroduced the new system?

17. Telephone area codes are three-digit numbers of theform XXX.

a. Originally, the first and third digits were neither 0 nor1 and the second digit was always a 0 or a 1. Howmany three-digit numbers of this type are possible?

b. Over time, the restrictions listed in part (a) havebeen altered; currently, the only requirement is thatthe first digit is neither 0 nor 1. How many three-digit numbers of this type are possible?

2.3 Exercises

100

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y Use the Fundamental Principle of Counting to

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y Use the Fundamental Principle of Counting todetermine how many different meal combinations

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y determine how many different meal combinations

Construct a tree diagram to list all possible soup,

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y Construct a tree diagram to list all possible soup,sandwich, and beverage combinations.

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y sandwich, and beverage combinations.

If you buy three pairs of jeans, four sweaters, and two

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y If you buy three pairs of jeans, four sweaters, and twopairs of boots, how many new outfits (consisting of a

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pairs of boots, how many new outfits (consisting of anew pair of jeans, a new sweater, and a new pair ofProp

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new pair of jeans, a new sweater, and a new pair ofboots) will you have?Prop

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boots) will you have?

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A certain model of automobile is available in six

of (minestrone and split pea), and three beverages (coffee,

of (minestrone and split pea), and three beverages (coffee,

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sandwiches (turkey, tuna, and tofu), two soups(minestrone and split pea), and three beverages (coffee,Cen

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(minestrone and split pea), and three beverages (coffee,

A cafeteria offers a complete dinner that includes one

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serving each of appetizer, soup, entrée, and dessert for

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$6.99. If the menu has three appetizers, four soups, six

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entrées, and three desserts, how many different meals

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entrées, and three desserts, how many different mealsare possible?

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are possible?

12.

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12. A sandwich shop offers a “U-Chooz” special consisting

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A sandwich shop offers a “U-Chooz” special consistingof your choice of bread, meat, cheese, and special sauce

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of your choice of bread, meat, cheese, and special sauce(one each). If there are six different breads, eight meats,

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(one each). If there are six different breads, eight meats,

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ing types of poles. Assuming that all items are compatible

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ing types of poles. Assuming that all items are compatible

with each other, how many different complete ski

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ing with each other, how many different complete ski

equipment packages are available?

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equipment packages are available?

An audio equipment store has ten different amplifiers,

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An audio equipment store has ten different amplifiers,four tuners, six turntables, eight tape decks, six compact

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four tuners, six turntables, eight tape decks, six compactdisc players, and thirteen speakers. Assuming that all

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disc players, and thirteen speakers. Assuming that allcomponents are compatible with each other, how many

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components are compatible with each other, how manydifferent complete stereo systems are available?

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different complete stereo systems are available?

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A cafeteria offers a complete dinner that includes oneserving each of appetizer, soup, entrée, and dessert forLe

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serving each of appetizer, soup, entrée, and dessert for

41. Find the value of when n � 5 and r � 5.

42. Find the value of when n � r.




46. Find the value of when n � r.



47. What is the Fundamental Principle of Counting? Whenis it used?

48. What is a factorial?


49. Who invented the modern symbol denoting a factorial?What symbol did it replace? Why?

Exercises 50–54 refer to the following: In an executive parkinglot, there are six parking spaces in a row, labeled 1 through 6.Exactly five cars of five different colors—black, gray, pink, white,and yellow—are to be parked in the spaces. The cars can park inany of the spaces as long as the following conditions are met:

● The pink car must be parked in space 3.● The black car must be parked in a space next to the space

in which the yellow car is parked.● The gray car cannot be parked in a space next to the space

in which the white car is parked.

50. If the yellow car is parked in space 1, how many accep-table parking arrangements are there for the five cars?

a. 1 b. 2 c. 3 d. 4 e. 5

51. Which of the following must be true of any acceptableparking arrangement?

a. One of the cars is parked in space 2.

b. One of the cars is parked in space 6.


n!1n � r 2 !r!

n!1n � r 2 !r!

n!1n � r 2 !r!

n!1n � r 2 !r!

n!1n � r 2 !

n!1n � r 2 !c. Why were the original restrictions listed in part (a)altered?

18. Major credit cards such as VISA and MasterCard havea sixteen-digit account number of the form XXXX-XXXX-XXXX-XXXX. How many different numbersof this type are possible?

19. The serial number on a dollar bill consists of a letterfollowed by eight digits and then a letter. How manydifferent serial numbers are possible, given thefollowing conditions?

a. Letters and digits cannot be repeated.

b. Letters and digits can be repeated.

c. The letters are nonrepeated consonants and thedigits can be repeated.

20. The serial number on a new twenty-dollar bill consistsof two letters followed by eight digits and then a letter.How many different serial numbers are possible, giventhe following conditions?

a. Letters and digits cannot be repeated.

b. Letters and digits can be repeated.

c. The first and last letters are repeatable vowels, thesecond letter is a consonant, and the digits can berepeated.

21. Each student at State University has a student I.D.number consisting of four digits (the first digit isnonzero, and digits may be repeated) followed by threeof the letters A, B, C, D, and E (letters may not berepeated). How many different student numbers arepossible?

22. Each student at State College has a student I.D.number consisting of five digits (the first digit isnonzero, and digits may be repeated) followed by twoof the letters A, B, C, D, and E (letters may not berepeated). How many different student numbers arepossible?

In Exercises 23–38, find the indicated value.

23. 4! 24. 5!

25. 10! 26. 8!

27. 20! 28. 25!

29. 6! · 4! 30. 8! · 6!

31. a. b. 32. a. b.

33. 34.

35. 36.

37. 38.


40. Find the value of when n � 19 and r � 16.n!1n � r 2 !

n!1n � r 2 !

77!

74! · 3!

82!

80! · 2!

6!

3! · 3!

8!

4! · 4!

9!

5! · 4!

8!

5! · 3!

8!

2!

8!

6!

6!

2!

6!

4!

2.3 Exercises 101

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y nonzero, and digits may be repeated) followed by two

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24.

of Each student at State College has a student I.D.

of Each student at State College has a student I.D.number consisting of five digits (the first digit isof number consisting of five digits (the first digit isnonzero, and digits may be repeated) followed by twoof nonzero, and digits may be repeated) followed by two

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What symbol did it replace? Why?

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What symbol did it replace? Why?

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Each student at State University has a student I.D.

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Each student at State University has a student I.D.number consisting of four digits (the first digit is

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number consisting of four digits (the first digit isnonzero, and digits may be repeated) followed by three

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nonzero, and digits may be repeated) followed by threeof the letters A, B, C, D, and E (letters may not be

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of the letters A, B, C, D, and E (letters may not berepeated). How many different student numbers areCen

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sentences and your own words.

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ing sentences and your own words.

Concept Questions

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What is the Fundamental Principle of Counting? When

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What is the Fundamental Principle of Counting? When

What is a factorial?

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What is a factorial?

History QuestionsLearn

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History Questions

Who invented the modern symbol denoting a factorial?Le

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Who invented the modern symbol denoting a factorial?


54. If the yellow car is parked in space 2, which of thefollowing must be true?

a. None of the cars is parked in space 5.

b. The gray car is parked in space 6.

c. The black car is parked in a space next to the spacein which the white car is parked.

d. The white car is parked in a space next to the spacein which the pink car is parked.

e. The gray car is parked in a space next to the space inwhich the black car is parked.

c. There is an empty space next to the space in whichthe gray car is parked.

d. There is an empty space next to the space in whichthe yellow car is parked.

e. Either the black car or the yellow car is parked in aspace next to space 3.

52. If the gray car is parked in space 2, none of the cars canbe parked in which space?

a. 1 b. 3 c. 4 d. 5 e. 6

53. The white car could be parked in any of the spacesexcept which of the following?

a. 1 b. 2 c. 4 d. 5 e. 6

2.4 Permutations and Combinations

Objectives

• Develop and apply the Permutation Formula

• Develop and apply the Combination Formula

• Determine the number of distinguishable permutations

The Fundamental Principle of Counting allows us to determine the total number ofpossible outcomes when a series of decisions (making selections from variouscategories) must be made. In Section 2.3, the examples and exercises involvedselecting one item each from various categories; if you buy two pairs of jeans,three shirts, and two pairs of shoes, you will have twelve (2 · 3 · 2 � 12) new out-fits (consisting of a new pair of jeans, a new shirt, and a new pair of shoes). In thissection, we examine the situation when more than one item is selected from a cat-egory. If more than one item is selected, the selections can be made either with orwithout replacement.

With versus Without Replacement

Selecting items with replacement means that the same item can be selected morethan once; after a specific item has been chosen, it is put back into the pool of fu-ture choices. Selecting items without replacement means that the same item cannotbe selected more than once; after a specific item has been chosen, it is not replaced.

Suppose you must select a four-digit Personal Identification Number (PIN)for a bank account. In this case, the digits are selected with replacement; eachtime a specific digit is selected, the digit is put back into the pool of choices forthe next selection. (Your PIN can be 3666; the same digit can be selected morethan once.) When items are selected with replacement, we use the FundamentalPrinciple of Counting to determine the total number of possible outcomes; thereare 10 · 10 · 10 · 10 � 10,000 possible four-digit PINs.

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of The Fundamental Principle of Counting allows us to determine the total number ofof The Fundamental Principle of Counting allows us to determine the total number ofpossible outcomes when a series of decisions (making selections from variousof possible outcomes when a series of decisions (making selections from variouscategories) must be made. In Section 2.3, the examples and exercises involvedof categories) must be made. In Section 2.3, the examples and exercises involved

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Develop and apply the Permutation Formula

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Develop and apply the Permutation Formula

Develop and apply the Combination Formula

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Develop and apply the Combination Formula

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Determine the number of distinguishable permutations

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Permutations and CombinationsLe

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Permutations and Combinations

In many situations, items cannot be selected more than once. For instance, whenselecting a committee of three people from a group of twenty, you cannot select thesame person more than once. Once you have selected a specific person (say, Lauren),you do not put her back into the pool of choices. When selecting items withoutreplacement, depending on whether the order of selection is important, permutationsor combinations are used to determine the total number of possible outcomes.

Permutations

When more than one item is selected (without replacement) from a single category,and the order of selection is important, the various possible outcomes are calledpermutations. For example, when the rankings (first, second, and third place) in a tal-ent contest are announced, the order of selection is important; Monte in first, Lynn insecond, and Ginny in third place is different from Ginny in first, Monte in second, andLynn in third. “Monte, Lynn, Ginny” and “Ginny, Monte, Lynn” are different permu-tations of the contestants. Naturally, these selections are made without replacement;we cannot select Monte for first place and reselect him for second place.

EXAMPLE 1 FINDING THE NUMBER OF PERMUTATIONS Six local bands havevolunteered to perform at a benefit concert, but there is enough time for only fourbands to play. There is also some concern over the order in which the chosen bandswill perform. How many different lineups are possible?

SOLUTION We must select four of the six bands and put them in a specific order. The bands areselected without replacement; a band cannot be selected to play and then be rese-lected to play again. Because we must make four decisions, we draw four boxesand put the number of choices for each decision in each appropriate box. There aresix choices for the opening band. Naturally, the opening band could not be the follow-up act, so there are only five choices for the next group. Similarly, there are fourcandidates for the third group and three choices for the closing band. The totalnumber of different lineups possible is found by multiplying the number of choicesfor each decision:

� � � � 360

opening closingband band

With four out of six bands playing in the performance, 360 lineups are possible.Because the order of selecting the bands is important, the various possible out-comes, or lineups, are called permutations; there are 360 permutations of six itemswhen the items are selected four at a time.

The computation in Example 1 is similar to a factorial, but the factors do notgo all the way down to 1; the product 6 · 5 · 4 · 3 is a “truncated” (cut-off) facto-rial. We can change this truncated factorial into a complete factorial in the follow-ing manner:

multiplying by and

�6!

2!

11

22

6 · 5 · 4 · 3 �6 · 5 · 4 · 3 · 12 · 1 2 12 · 1 2

↑↑3456

2.4 Permutations and Combinations 103

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volunteered to perform at a benefit concert, but there is enough time for only four

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bands to play. There is also some concern over the order in which the chosen bandswill perform. How many different lineups are possible?

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will perform. How many different lineups are possible?

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ing When more than one item is selected (without replacement) from a single category,

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ing When more than one item is selected (without replacement) from a single category,

important, the various possible outcomes are called

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ing important, the various possible outcomes are called

For example, when the rankings (first, second, and third place) in a tal-

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For example, when the rankings (first, second, and third place) in a tal-ent contest are announced, the order of selection is important; Monte in first, Lynn in

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ent contest are announced, the order of selection is important; Monte in first, Lynn insecond, and Ginny in third place is different from Ginny in first, Monte in second, and

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second, and Ginny in third place is different from Ginny in first, Monte in second, andLynn in third. “Monte, Lynn, Ginny” and “Ginny, Monte, Lynn” are different permu-

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Lynn in third. “Monte, Lynn, Ginny” and “Ginny, Monte, Lynn” are different permu-tations of the contestants. Naturally, these selections are made without replacement;

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tations of the contestants. Naturally, these selections are made without replacement;we cannot select Monte for first place and reselect him for second place.

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we cannot select Monte for first place and reselect him for second place.

FINDING THE NUMBER OF PERMUTATIONSLearn

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FINDING THE NUMBER OF PERMUTATIONS

Notice that this last expression can be written as . (Recall that we wereselecting four out of six bands.) This result is generalized as follows.

6!2! � 6!16 � 4 2 !


Using the notation above and referring to Example 1, we note that 360 pos-sible lineups of four bands selected from a pool of six can be denoted by

Other notations can be used to represent the number of per-mutations of a group of items. In particular, the notations nPr, P(n, r), Pr

n, and Pn, r

all represent the number of possible permutations (or arrangements) of r itemsselected (without replacement) from a pool of n items.

EXAMPLE 2 FINDING THE NUMBER OF PERMUTATIONS Three door prizes (first,second, and third) are to be awarded at a ten-year high school reunion. Each of the112 attendees puts his or her name in a hat. The first name drawn wins a two-nightstay at the Chat ’n’ Rest Motel, the second name wins dinner for two at Juju’sKitsch-Inn, and the third wins a pair of engraved mugs. In how many differentways can the prizes be awarded?

SOLUTION We must select 3 out of 112 people (without replacement), and the order inwhich they are selected is important. (Winning dinner is different from winning themugs.) Hence, we must find the number of permutations of 3 items selected froma pool of 112:

There are 1,367,520 different ways in which the three prizes can be awarded to the112 people.

In Example 2, if you try to use a calculator to find directly, you will notobtain an answer. Entering 112 and pressing results in a calculator error. (Tryit.) Because factorials get very large very quickly, most calculators are not able tofind any factorial over 69!. (69! � 1.711224524 � 1098.)

x!

112!109!

� 1,367,520 � 112 # 111 # 110

�112 · 111 · 110 · 109 · 108 · · · · · 2 · 1 109 · 108 · · · · · 2 · 1

�112!

109!

112P3 �112!

1112 � 3 2 !

6P4 � 6!16 � 4 2 ! � 360.

PERMUTATION FORMULAThe number of permutations, or arrangements, of r items selected withoutreplacement from a pool of n items (r � n), denoted by nPr, is

Permutations are used whenever more than one item is selected (withoutreplacement) from a category and the order of selection is important.

nPr �n!

1n � r 2 !

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Using the notation above and referring to Example 1, we note that 360 pos-sible lineups of four bands selected from a pool of six can be denoted by

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EXAMPLE 3 FINDING THE NUMBER OF PERMUTATIONS A bowling league has tenteams. In how many different ways can the teams be ranked in the standings at theend of a tournament? (Ties are not allowed.)

SOLUTION Because order is important, we find the number of permutations of ten items selectedfrom a pool of ten items:

Recall that 0! � 1.

In a league containing ten teams, 3,628,800 different standings are possible at theend of a tournament.

Combinations

When items are selected from a group, the order of selection may or may not beimportant. If the order is important (as in Examples 1, 2, and 3), permutations areused to determine the total number of selections possible. What if the order ofselection is not important? When more than one item is selected (without replace-ment) from a single category and the order of selection is not important, the vari-ous possible outcomes are called combinations.

EXAMPLE 4 LISTING ALL POSSIBLE COMBINATIONS Two adults are needed tochaperone a daycare center’s field trip. Marcus, Vivian, Frank, and Keiko are thefour managers of the center. How many different groups of chaperones arepossible?

SOLUTION In selecting the chaperones, the order of selection is not important; “Marcus andVivian” is the same as “Vivian and Marcus.” Hence, the permutation formulacannot be used. Because we do not yet have a shortcut for finding the total numberof possibilities when the order of selection is not important, we must list all thepossibilities:

Marcus and Vivian Marcus and Frank Marcus and Keiko

Vivian and Frank Vivian and Keiko Frank and Keiko

Therefore, six different groups of two chaperones are possible from the groupof four managers. Because the order in which the people are selected is notimportant, the various possible outcomes, or groups of chaperones, are calledcombinations; there are six combinations when two items are selected from apool of four.

Just as nPr denotes the number of permutations of r elements selected from apool of n elements, nCr denotes the number of combinations of r elements selectedfrom a pool of n elements. In Example 4, we found that there are six combinations of

� 3,628,800

�10!

1

�10!

0!

10P10 �10!

110 � 10 2 !


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y LISTING ALL POSSIBLE COMBINATIONS

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y LISTING ALL POSSIBLE COMBINATIONSchaperone a daycare center’s field trip. Marcus, Vivian, Frank, and Keiko are the

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y chaperone a daycare center’s field trip. Marcus, Vivian, Frank, and Keiko are thefour managers of the center. How many different groups of chaperones are

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y four managers of the center. How many different groups of chaperones arepossible?

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y possible?

SOLUTION

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SOLUTION

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y of

ous possible outcomes are called

of ous possible outcomes are called

LISTING ALL POSSIBLE COMBINATIONSof LISTING ALL POSSIBLE COMBINATIONS

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When items are selected from a group, the order of selection may or may not be

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When items are selected from a group, the order of selection may or may not beimportant. If the order is important (as in Examples 1, 2, and 3), permutations are

Cenga

ge

important. If the order is important (as in Examples 1, 2, and 3), permutations areused to determine the total number of selections possible. What if the order of

Cenga

ge

used to determine the total number of selections possible. What if the order ofnot

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not important? When more than one item is selected (without replace-

Cenga

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important? When more than one item is selected (without replace-ment) from a single category and the order of selection is not important, the vari-

Cenga

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ment) from a single category and the order of selection is not important, the vari-ous possible outcomes are called Cen

gage

ous possible outcomes are called

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In a league containing ten teams, 3,628,800 different standings are possible at the

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ing

In a league containing ten teams, 3,628,800 different standings are possible at the

two people selected from a pool of four by listing all six of the combinations; that is,4C2 � 6. If we had a larger pool, listing each combination to find out how many thereare would be extremely time consuming and tedious! Instead of listing, we take a dif-ferent approach. We first find the number of permutations (with the permutation for-mula) and then alter that number to account for the distinction between permutationsand combinations.

To find the number of combinations of two people selected from a pool offour, we first find the number of permutations:

This figure of 12 must be altered to account for the distinction between permuta-tions and combinations.

In Example 4, we listed combinations; one such combination was “Marcusand Vivian.” If we had listed permutations, we would have had to list both“Marcus and Vivian” and “Vivian and Marcus,” because the order of selectionmatters with permutations. In fact, each combination of two chaperones listed inExample 4 generates two permutations; each pair of chaperones can be given intwo different orders. Thus, there are twice as many permutations of two people se-lected from a pool of four as there are combinations. Alternatively, there are half asmany combinations of two people selected from a pool of four as there are permu-tations. We used the permutation formula to find that 4P2 � 12; thus,

This answer certainly fits with Example 4; we listed exactly six combinations.What if three of the four managers were needed to chaperone the daycare

center’s field trip? Rather than finding the number of combinations by listing eachpossibility, we first find the number of permutations and then alter that number toaccount for the distinction between permutations and combinations.

The number of permutations of three people selected from a pool of four is

We know that some of these permutations represent the same combination. Forexample, the combination “Marcus and Vivian and Keiko” generates 3! � 6 dif-ferent permutations (using initials, they are: MVK, MKV, KMV, KVM, VMK,VKM). Because each combination of three people generates six different permuta-tions, there are one-sixth as many combinations as permutations. Thus,

This means that if three of the four managers were needed to chaperone the day-care center’s field trip, there would be 4C3 � 4 possible combinations.

We just saw that when two items are selected from a pool of n items, eachcombination of two generates 2! � 2 permutations, so

We also saw that when three items are selected from a pool of n items, each com-bination of three generates 3! � 6 permutations, so

nC3 �1

3! · nP3

nC2 �1

2! · nP2

4C3 �1

6 · 4P3 �

1

6 124 2 � 4

4P3 �4!

14 � 3 2 ! �4!

1!� 24

4C2 �1

2 · 4P2 �

1

2 112 2 � 6

4P2 �4!

14 � 2 2 ! �4!

2!� 12


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y We know that some of these permutations represent the same combination. For

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y We know that some of these permutations represent the same combination. Forexample, the combination “Marcus and Vivian and Keiko” generates 3!

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y example, the combination “Marcus and Vivian and Keiko” generates 3!ferent permutations (using initials, they are: MVK, MKV, KMV, KVM, VMK,

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y ferent permutations (using initials, they are: MVK, MKV, KMV, KVM, VMK,VKM). Because each combination of three people generates six different permuta-

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VKM). Because each combination of three people generates six different permuta-

4

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possibility, we first find the number of permutations and then alter that number to

of possibility, we first find the number of permutations and then alter that number toaccount for the distinction between permutations and combinations.

of account for the distinction between permutations and combinations.

The number of permutations of three people selected from a pool of four isof The number of permutations of three people selected from a pool of four is

4!of

4!

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ge lected from a pool of four as there are combinations. Alternatively, there are half as

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ge lected from a pool of four as there are combinations. Alternatively, there are half as

many combinations of two people selected from a pool of four as there are permu-

Cenga

ge many combinations of two people selected from a pool of four as there are permu-

tations. We used the permutation formula to find that

Cenga

ge tations. We used the permutation formula to find that

This answer certainly fits with Example 4; we listed exactly six combinations.

Cenga

ge

This answer certainly fits with Example 4; we listed exactly six combinations.What if three of the four managers were needed to chaperone the daycare

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What if three of the four managers were needed to chaperone the daycarecenter’s field trip? Rather than finding the number of combinations by listing eachCen

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center’s field trip? Rather than finding the number of combinations by listing eachpossibility, we first find the number of permutations and then alter that number toCen

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possibility, we first find the number of permutations and then alter that number toaccount for the distinction between permutations and combinations.Cen

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account for the distinction between permutations and combinations.

�

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� 6

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6

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ing This figure of 12 must be altered to account for the distinction between permuta-

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ing This figure of 12 must be altered to account for the distinction between permuta-

In Example 4, we listed combinations; one such combination was “Marcus

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ing

In Example 4, we listed combinations; one such combination was “Marcusand Vivian.” If we had listed permutations, we would have had to list both

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ing

and Vivian.” If we had listed permutations, we would have had to list bothcus and Vivian” and “Vivian and Marcus,” because the

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ing

cus and Vivian” and “Vivian and Marcus,” because the order

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ordermatters with permutations. In fact, each combination of two chaperones listed in

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ing

matters with permutations. In fact, each combination of two chaperones listed inExample 4 generates two permutations; each pair of chaperones can be given in

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ing

Example 4 generates two permutations; each pair of chaperones can be given intwo different orders. Thus, there are twice as many permutations of two people se-

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two different orders. Thus, there are twice as many permutations of two people se-lected from a pool of four as there are combinations. Alternatively, there are half asLe

arning

lected from a pool of four as there are combinations. Alternatively, there are half asmany combinations of two people selected from a pool of four as there are permu-Le

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many combinations of two people selected from a pool of four as there are permu-tations. We used the permutation formula to find that

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ing

tations. We used the permutation formula to find that

More generally, when r items are selected from a pool of n items, each combina-tion of r items generates r! permutations, so

using the Permutation Formula

multiplying the fractions together �n!

r! · 1n � r 2 !

�1

r! ·

n!

1n � r 2 !

nCr �1

r! · nPr


EXAMPLE 5 FINDING THE NUMBER OF COMBINATIONS A DVD club sends you abrochure that offers any five DVDs from a group of fifty of today’s hottest releases.How many different selections can you make?

SOLUTION Because the order of selection is not important, we find the number of combinationswhen five items are selected from a pool of fifty:

� 2,118,760

�50 · 49 · 48 · 47 · 46

5 · 4 · 3 · 2 · 1

�50!

45! 5!

50C5 �50!

150 � 5 2 ! 5!

Graphing calculators have buttons that will calculate nPr and nCr. To use them,do the following:

• Type the value of n. (For example, type the number 50.)• Press the button.• Press the right arrow button as many times as necessary to highlight .• Press the down arrow button as many times as necessary to highlight the appropriate

symbol— for permutations, for combinations—and press .• Type the value of r. (For example, type the number 5.)• Press to execute the calculation.ENTER

ENTERnCrnPr

T

PRBS

MATH

50 45 5 �)x!�x!(�x!

COMBINATION FORMULAThe number of distinct combinations of r items selected without replacementfrom a pool of n items (r � n), denoted by nCr, is

Combinations are used whenever one or more items are selected (withoutreplacement) from a category and the order of selection is not important.

nCr �n!

1n � r 2 ! r!

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y of when five items are selected from a pool of fifty:

of when five items are selected from a pool of fifty:

50of 50Cof C5 of 5 C5 Cof C5 C

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FINDING THE NUMBER OF COMBINATIONS

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FINDING THE NUMBER OF COMBINATIONSbrochure that offers any five DVDs from a group of fifty of today’s hottest releases.

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brochure that offers any five DVDs from a group of fifty of today’s hottest releases.How many different selections can you make?

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How many different selections can you make?

Because the order of selection is Cenga

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Because the order of selection is when five items are selected from a pool of fifty:Cen

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when five items are selected from a pool of fifty:Cenga

ge replacement) from a category and the order of selection is not important.

Cenga

ge replacement) from a category and the order of selection is not important.

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items selected without replacement

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items selected without replacementn

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nC

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Cr

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r, is

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, is

Combinations are used whenever one or more items are selected (withoutLearn

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Combinations are used whenever one or more items are selected (withoutreplacement) from a category and the order of selection is not important.Le

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replacement) from a category and the order of selection is not important.Learn

ing

On a Casio graphing calculator, do the following:

• Press the button; this gives you access to the main menu.• Press 1 to select the RUN mode; this mode is used to perform arithmetic operations.• Type the value of n. (For example, type the number 50.)• Press the button; this gives you access to various options displayed at the bot-

tom of the screen.• Press the button to see more options (i.e., ).• Press the button to select probability options (i.e., ).• Press the button to select combinations (i.e., ) or the button to select per-

mutations (i.e., ).• Type the value of r. (For example, type the number 5.)• Press the button to execute the calculation.EXE

nPr

F2nCrF3PROBF3

SF6

OPTN

MENU


In choosing five out of fifty DVDs, 2,118,760 combinations are possible.

EXAMPLE 6 FINDING THE NUMBER OF COMBINATIONS A group consisting oftwelve women and nine men must select a five-person committee. How manydifferent committees are possible if it must consist of the following?

a. three women and two men b. any mixture of men and women

SOLUTION a. Our problem involves two categories: women and men. The Fundamental Principle ofCounting tells us to draw two boxes (one for each category), enter the number of choicesfor each, and multiply:

� � ?

Because the order of selecting the members of a committee is not important, we will usecombinations:

� 220 · 36

� 7,920

�12 · 11 · 10

3 · 2 · 1 ·

9 · 82 · 1

�12!

9! · 3! ·

9!

7! · 2!

112C3 2 · 19C2 2 � 12!

112 � 3 2 ! · 3! ·

9!

19 � 2 2 ! · 2!

the number of waysin which we canselect two out

of nine men

the number of waysin which we canselect three out

of twelve women

50 5

Casio 50 (i.e., ) (i.e., ) (i.e., ) 5

EXE

F3nCrF3PROBF6SOPTN

ENTERnCrPRBMATH

95057_02_ch02_p067-130.qxd 9/27/10 9:53 AM Page 108

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y Our problem involves two categories: women and men. The Fundamental Principle of

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y Our problem involves two categories: women and men. The Fundamental Principle ofCounting tells us to draw two boxes (one for each category), enter the number of choices

Propert

y Counting tells us to draw two boxes (one for each category), enter the number of choicesfor each, and multiply:

Propert

y for each, and multiply:

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y of different committees are possible if it must consist of the following?

of different committees are possible if it must consist of the following?

three women and two menof three women and two men

Our problem involves two categories: women and men. The Fundamental Principle ofof Our problem involves two categories: women and men. The Fundamental Principle of

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FINDING THE NUMBER OF COMBINATIONStwelve women and nine men must select a five-person committee. How manyCen

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twelve women and nine men must select a five-person committee. How manydifferent committees are possible if it must consist of the following?Cen

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different committees are possible if it must consist of the following?

three women and two menCen

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three women and two menCen

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(i.e., ) (i.e., ) (i.e., ) 5

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(i.e., ) (i.e., ) (i.e., ) 5

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F3

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F3

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PROB

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PROB

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There are 7,920 different committees consisting of three women and two men.b. Because the gender of the committee members doesn’t matter, our problem involves

only one category: people. We must choose five out of the twenty-one people, and theorder of selection is not important:

� 20,349

�21 · 20 · 19 · 18 · 17

5 · 4 · 3 · 2 · 1

�21!

16! · 5!

21C5 �21!

121 � 5 2 ! · 5!


There are 20,349 different committees consisting of five people.

EXAMPLE 7 EVALUATING THE COMBINATION FORMULA Find the value of 5Cr forthe following values of r:

a. r � 0 b. r � 1 c. r � 2 d. r � 3 e. r � 4 f. r � 5

SOLUTION a.

b.

c.

d.

e.

f. 5C5 �5!

15 � 5 2 ! · 5!�

5!

0! · 5!� 1

5C4 �5!

15 � 4 2 ! · 4!�

5!

1! · 4!� 5

5C3 �5!

15 � 3 2 ! · 3!�

5!

2! · 3! � 10

5C2 �5!

15 � 2 2 ! · 2!�

5!

3! · 2!� 10

5C1 �5!

15 � 1 2 ! · 1!�

5!

4! · 1!� 5

5C0 �5!

15 � 0 2 ! · 0!�

5!

5! · 0!� 1

12 9 3 9

7 2

12 3 9 2 ENTERnCrPRBMATH�nCrPRBMATH

�))x!�x!(

�x!(�))x!�x!(�x!(

21 16 5

21 5 ENTERnCrPRBMATH

�)x!�x!(�x!

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y There are 20,349 different committees consisting of five people.

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y There are 20,349 different committees consisting of five people.

EXAMPLE

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SOLUTIONProp

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SOLUTIONProp

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16 5

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16 5

21 Cenga

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21 Cenga

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MATHCenga

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MATHCenga

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x

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x!

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!

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(

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(

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only one category: people. We must choose five out of the twenty-one people, and the

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ing only one category: people. We must choose five out of the twenty-one people, and the

The combinations generated in Example 7 exhibit a curious pattern. Noticethat the values of 5Cr are symmetric: 5C0 � 5C5, 5C1 � 5C4, and 5C2 � 5C3. Nowexamine the diagram in Figure 2.41. Each number in this “triangle” of numbers isthe sum of two numbers in the row immediately above it. For example, 2 � 1 � 1and 10 � 4 � 6, as shown by the inserted arrows. It is no coincidence that the val-ues of 5Cr found in Example 7 also appear as a row of numbers in this “magic”triangle. In fact, the sixth row contains all the values of 5Cr for r � 0, 1, 2, 3, 4,and 5. In general, the (n � 1)th row of the triangle contains all the values ofnCr for r � 0, 1, 2, . . . , n; alternatively, the nth row of the triangle contains allthe values of n�1Cr for r � 0, 1, 2, . . . , n�1. For example, the values of 9Cr, forr � 0, 1, 2, . . . , 9, are in the tenth row, and vice versa, the entries in the tenth roware the values of 9Cr, for r � 0, 1, 2, . . . , 9.

Historically, this triangular pattern of numbers is referred to as Pascal’s Tri-angle, in honor of the French mathematician, scientist, and philosopher BlaisePascal (1623–1662). Pascal is a cofounder of probability theory (see the HistoricalNote in Section 3.1). Although the triangle has Pascal’s name attached to it, this“magic” arrangement of numbers was known to other cultures hundreds of yearsbefore Pascal’s time.

The most important part of any problem involving combinatorics is decidingwhich counting technique (or techniques) to use. The following list of generalsteps and the flowchart in Figure 2.42 can help you to decide which method ormethods to use in a specific problem.


Pascal’s triangle.FIGURE 2.41

1st row

2nd row

5th row

6th row

and so on

1 1

11

1 1

11

1 1

1

2

3 3

44 6

5 510 10

3rd row

4th row

WHICH COUNTING TECHNIQUE?1. What is being selected?2. If the selected items can be repeated, use the Fundamental Principle of

Counting and multiply the number of choices for each category.3. If there is only one category, use:

combinations if the order of selection does not matter—that is, r items can beselected from a pool of n items in nCr � ways.permutations if the order of selection does matter—that is, r items can beselected from a pool of n items in nPr � ways.

4. If there is more than one category, use the Fundamental Principle of Countingwith one box per category.a. If you are selecting one item per category, the number in the box for that

category is the number of choices for that category.b. If you are selecting more than one item per category, the number in the box

for that category is found by using step 3.

n!1n � r 2 !

n!1n � r 2 ! · r!

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y “magic” arrangement of numbers was known to other cultures hundreds of years

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y “magic” arrangement of numbers was known to other cultures hundreds of yearsbefore Pascal’s time.

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y before Pascal’s time.The most important part of any problem involving combinatorics is deciding

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y The most important part of any problem involving combinatorics is decidingwhich counting technique (or techniques) to use. The following list of general

Propert

y which counting technique (or techniques) to use. The following list of generalsteps and the flowchart in Figure 2.42 can help you to decide which method or

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y steps and the flowchart in Figure 2.42 can help you to decide which method ormethods to use in a specific problem.

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y methods to use in a specific problem.

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y

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y of

in honor of the French mathematician, scientist, and philosopher Blaise

of in honor of the French mathematician, scientist, and philosopher Blaise

Pascal (1623–1662). Pascal is a cofounder of probability theory (see the Historical

of Pascal (1623–1662). Pascal is a cofounder of probability theory (see the HistoricalNote in Section 3.1). Although the triangle has Pascal’s name attached to it, thisof Note in Section 3.1). Although the triangle has Pascal’s name attached to it, this“magic” arrangement of numbers was known to other cultures hundreds of yearsof “magic” arrangement of numbers was known to other cultures hundreds of years

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Historically, this triangular pattern of numbers is referred to as Cenga

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Historically, this triangular pattern of numbers is referred to as in honor of the French mathematician, scientist, and philosopher BlaiseCen

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in honor of the French mathematician, scientist, and philosopher BlaisePascal (1623–1662). Pascal is a cofounder of probability theory (see the HistoricalCen

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Pascal (1623–1662). Pascal is a cofounder of probability theory (see the Historical

Pascal’s triangle.

Cenga

ge

Pascal’s triangle.

1 1

Cenga

ge

1 1

1

Cenga

ge 1

5 5

Cenga

ge

5 51 15 51 1

Cenga

ge

1 15 51 1

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ing


What isbeing selected?

Can the selecteditems be repeated?

Is there more thanone category?

Use the FundamentalPrinciple of Counting.

Does the order ofselection matter?

Usepermutations.

Usecombinations.

Usecombinations.

Usepermutations.

Does the order ofselection matter?

Am I selecting onlyone item per category?

Enter the number of choicesfor that category in its box.

Use the FundamentalPrinciple of Counting,with one box per category.For each box, ask thefollowing question:

No

No

No

No

No

Yes

Yes

Yes Yes

Yes

Start

Which counting technique?FIGURE 2.42

EXAMPLE 8 USING THE “WHICH COUNTING TECHNIQUE?” FLOWCHART Astandard deck of playing cards contains fifty-two cards.

a. How many different five-card hands containing four kings are possible?b. How many different five-card hands containing four queens are possible?c. How many different five-card hands containing four kings or four queens are possible?d. How many different five-card hands containing four of a kind are possible?

SOLUTION a. We use the flowchart in Figure 2.42 and answer the following questions.

Q. What is being selected?A. Playing cards.

Q. Can the selected items be repeated?A. No.

Q. Is there more than one category?A. Yes: Because we must have five cards, we need four kings and one non-king.

Therefore, we need two boxes:

�

Q. Am I selecting only one item per category?A. Kings: no. Does the order of selection matter? No: Use combinations. Because

there are n � 4 kings in the deck and we want to select r � 4, we must compute

4C4. Non-kings: yes. Enter the number of choices for that category: There are48 non-kings.

non-kingskings

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y USING THE “WHICH COUNTING TECHNIQUE?” FLOWCHART

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y USING THE “WHICH COUNTING TECHNIQUE?” FLOWCHARTstandard deck of playing cards contains fifty-two cards.

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y standard deck of playing cards contains fifty-two cards.

a.

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y a. How many different five-card hands containing four kings are possible?

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y How many different five-card hands containing four kings are possible?

b.

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y b.

SOLUTIONPropert

y

SOLUTIONPropert

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USING THE “WHICH COUNTING TECHNIQUE?” FLOWCHARTof USING THE “WHICH COUNTING TECHNIQUE?” FLOWCHART

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ge

Cenga

ge

Cenga

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combinations.

Cenga

ge combinations.

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ge

Which counting technique

Cenga

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Which counting technique

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Enter the number of choices

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Enter the number of choicesfor that category in its box.

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for that category in its box.


� � �

There are forty-eight different five-card hands containing four kings.b. Using the same method as in part (a), we would find that there are forty-eight different

five-card hands containing four queens; the number of five-card hands containing fourqueens is the same as the number of five-card hands containing four kings.

c. To find the number of five-card hands containing four kings or four queens, we definethe following sets:

A � {five-card hands � the hand contains four kings}

B � {five-card hands � the hand contains four queens}

Consequently,

A ´ B � {five-card hands � the hand contains four kings or four queens}

A ¨ B � {five-card hands � the hand contains four kings and four queens}

(Recall that the union symbol, ´, may be interpreted as the word “or,” while theintersection symbol, ¨, may be interpreted as the word “and.” See Figure 2.13 for acomparison of set theory and logic.)

Because there are no five-card hands that contain four kings and four queens, wenote that n(A ¨ B) � 0.

� 48

� 1 · 48

�4!

0! · 4! · 48

�4!

14 � 4 2 ! · 4! · 48

484C4non-kingskings

Chu Shih-chieh was the last and mostacclaimed mathematician of the

Sung Dynasty in China. Little is knownof his personal life; the actual dates ofhis birth and death are unknown. Hiswork appears to have flourished duringthe close of the thirteenth century. It isbelieved that Chu Shih-chieh spentmany years as a wandering scholar,earning a living by teaching mathemat-ics to those who wanted to learn.

Two of Chu Shih-chieh’s works havesurvived the centuries. The first, Suan-hsüeh ch’i-meng (Introduction to Mathe-matical Studies), was written in 1299and contains elementary mathematics.This work was very influential in Japan

and Korea, although it was lost in Chinauntil the nineteenth century. Written in1303, Chu’s second work Ssu-yüan yü-chien (Precious Mirror of the Four Ele-ments) contains more advanced mathe-matics. The topics of Precious Mirrorinclude the solving of simultaneousequations and the solving of equationsup to the fourteenth degree.

Of the many diagrams in PreciousMirror, one has special interest: the arith-metic triangle. Chu Shih-chieh’s trianglecontains the first eight rows of what isknown in the West as Pascal’s Triangle.However, Chu does not claim credit forthe triangle; he refers to it as “a diagramof the old method for finding eighth andlower powers.” “Pascal’s” Triangle wasknown to the Chinese well over300 years before Pascal was born!

HistoricalNote

CHU SHIH-CHIEH, CIRCA 1280–1303

The “Pascal” Triangle as depicted in1303 at the front of Chu Shih-chieh’s Ssu-yüan yü-chien. It is entitled “The OldMethod Chart of the Seven MultiplyingSquares” and tabulates the binomialcoefficients up to the eighth power.

Brow

n Un

iver

sity

Lib

rary

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y There are forty-eight different five-card hands containing four kings.

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y There are forty-eight different five-card hands containing four kings.

b.

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y b. Using the same method as in part (a), we would find that there are forty-eight different

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y Using the same method as in part (a), we would find that there are forty-eight different

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y of C

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�

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� 4

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4C

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C

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non-kings

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non-kings

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known to the Chinese well over

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300 years before Pascal was born!

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300 years before Pascal was born!

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The “Pascal” Triangle as depicted inLearn

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The “Pascal” Triangle as depicted in1303 at the front of Chu Shih-chieh’s Learn

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1303 at the front of Chu Shih-chieh’s Ssu-yüan yü-chienLearn

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Ssu-yüan yü-chienMethod Chart of the Seven MultiplyingLearn

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Method Chart of the Seven Multiplying


Using the Union/Intersection Formula for the Union of Sets, we obtain

n(A ´ B) � n(A) � n(B) � n(A ¨ B)

� 48 � 48 � 0

� 96

There are ninety-six different five-card hands containing four kings or four queens.d. Four of a kind means four cards of the same “denomination,” that is, four 2’s, or four 3’s,

or four 4’s, or . . . , or four kings, or four aces. Now, regardless of the denomination of thecard, there are forty-eight different five-card hands that contain four of any specificdenomination; there are forty-eight different five-card hands that contain four 2’s, thereare forty-eight different five-card hands that contain four 3’s, there are forty-eight differentfive-card hands that contain four 4’s, and so on. As is shown in part (c), the word “or”implies that we add cardinal numbers. Consequently,

n(four of a kind) � n(four 2’s or four 3’s or . . . or four kings or four aces)

� n(four 2’s) � n(four 3’s � . . . � n(four kings) � n(four aces)

� 48 � 48 � � � � � 48 � 48 (thirteen times)

� 13 � 48

� 624

There are 624 different five-card hands containing four of a kind.

As is shown in Example 8, there are 624 possible five-card hands that containfour of a kind. When you are dealt five cards, what is the likelihood (or probabil-ity) that you will receive one of these hands? This question, and its answer, will beexplored in Section 3.4, “Combinatorics and Probability.”

Permutations of Identical Items

In how many different ways can the three letters in the word “SAW” be arranged?As we know, arrangements are referred to as permutations, so we can apply thePermutation Formula,

Therefore,

The six permutations of the letters in SAW are

SAW SWA AWS ASW WAS WSA

In general, if we have three different items (the letters in SAW), we can arrangethem in 3! � 6 ways. However, this method applies only if the items are alldifferent (distinct).

What happens if some of the items are the same (identical)? For example, in howmany different ways can the three letters in the word “SEE” be arranged? Because twoof the letters are identical (E), we cannot use the Permutation Formula directly; we takea slightly different approach. Temporarily, let us assume that the E’s are written in dif-ferent colored inks, say, red and blue. Therefore, SEE could be expressed as SEE.These three symbols could be arranged in 3! � 6 ways as follows:

SEE SEE ESE ESE EES EES

If we now remove the color, the arrangements are

SEE SEE ESE ESE EES EES

3P3 �3!

13 � 3 2 ! �3!

0!�

3 �2 �1

1� 6

nPr � n!1n � r 2 !.

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y In how many different ways can the three letters in the word “SAW” be arranged?

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of Permutations of Identical Items

In how many different ways can the three letters in the word “SAW” be arranged?of In how many different ways can the three letters in the word “SAW” be arranged?

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ge There are 624 different five-card hands containing four of a kind.

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ge There are 624 different five-card hands containing four of a kind.

As is shown in Example 8, there are 624 possible five-card hands that contain

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As is shown in Example 8, there are 624 possible five-card hands that containfour of a kind. When you are dealt five cards, what is the likelihood (or probabil-

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four of a kind. When you are dealt five cards, what is the likelihood (or probabil-ity) that you will receive one of these hands? This question, and its answer, will be

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ity) that you will receive one of these hands? This question, and its answer, will beexplored in Section 3.4, “Combinatorics and Probability.”

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explored in Section 3.4, “Combinatorics and Probability.”

Permutations of Identical ItemsCenga

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Permutations of Identical Items

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(four 2’s or four 3’s or . . . or four kings or four aces)

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(four 2’s or four 3’s or . . . or four kings or four aces)

(four 3’s

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(four 3’s �

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� . . .

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. . . �

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� n

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n(four kings)

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(four kings)

� � � � �

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� � � � � 48

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48 �

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� 48 (thirteen times)

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48 (thirteen times)

There are 624 different five-card hands containing four of a kind.Learn

ing

There are 624 different five-card hands containing four of a kind.

Some of these arrangements are duplicates of others; as we can see, there are onlythree different or distinguishable permutations, namely, SEE, ESE, and EES.Notice that when n � 3 (the total number of letters in SEE) and x � 2 (the numberof identical letters), we can divide n! by x! to obtain the number of distinguishablepermutations; that is,

This method is applicable because dividing by the factorial of the repeated lettereliminates the duplicate arrangements; the method may by generalized as follows.

n!

x!�

3!

2!�

3�2� 1

2 � 1�

3 � 2 � 1

2 � 1� 3


EXAMPLE 9 FINDING THE NUMBER OF DISTINGUISHABLE PERMUTATIONSFind the number of distinguishable permutations of the letters in the word“MISSISSIPPI.”

SOLUTION The word “MISSISSIPPI” has n � 11 letters; I is repeated x � 4 times, S is repeatedy � 4 times, and P is repeated z � 2 times. Therefore, we divide the total factorialby the factorial of each repeated letter and obtain

The letters in the word MISSISSIPPI can be arranged in 34,650 ways. (Note that ifthe 11 letters were all different, there would be 11! � 39,96,800 permutations.)

n!

x!y!z!�

11!

4!4!2!� 34,650

In Exercises 1–12, find the indicated value:

1. a. 7P3 b. 7C3 2. a. 8P4 b. 8C4

3. a. 5P5 b. 5C5 4. a. 9P0 b. 9C0

5. a. 14P1 b. 14C1 6. a. 13C3 b. 13C10

7. a. 100P3 b. 100C3 8. a. 80P4 b. 80C4

9. a. xPx�1 b. xCx�1 10. a. xP1 b. xC1

11. a. xP2 b. xC2 12. a. xPx�2 b. xCx�2

13. a. Find 3P2.

b. List all of the permutations of {a, b, c} when the elements are taken two at a time.

14. a. Find 3C2.

b. List all of the combinations of {a, b, c} when theelements are taken two at a time.

15. a. Find 4C2.

b. List all of the combinations of {a, b, c, d} when theelements are taken two at a time.

16. a. Find 4P2.

b. List all of the permutations of {a, b, c, d} when theelements are taken two at a time.

17. An art class consists of eleven students. All of themmust present their portfolios and explain their work tothe instructor and their classmates at the end of thesemester.

a. If their names are drawn from a hat to determinewho goes first, second, and so on, how many pre-sentation orders are possible?

2.4 Exercises

DISTINGUISHABLE PERMUTATIONS OF IDENTICAL ITEMSThe number of distinguishable permutations (or arrangements) of n itemsin which x items are identical, y items are identical, z items are identical, andso on, is . That is, to find the number of distinguishable permutations,divide the total factorial by the factorial of each repeated item.

n!x!y!z!p

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In Exercises 1–12, find the indicated value:Prop

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of The letters in the word MISSISSIPPI can be arranged in 34,650 ways. (Note that ifof The letters in the word MISSISSIPPI can be arranged in 34,650 ways. (Note that ifthe 11 letters were all different, there would be 11! of the 11 letters were all different, there would be 11!

4!4!2!

of 4!4!2!Cen

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FINDING THE NUMBER OF DISTINGUISHABLE PERMUTATIONS

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Find the number of distinguishable permutations of the letters in the word

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ge Find the number of distinguishable permutations of the letters in the word

The word “MISSISSIPPI” has

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The word “MISSISSIPPI” has n

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n �

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� 11 letters; I is repeated

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11 letters; I is repeated 4 times, and P is repeated

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4 times, and P is repeated z

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z �

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� 2 times. Therefore, we divide the total factorial

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2 times. Therefore, we divide the total factorialby the factorial of each repeated letter and obtain

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by the factorial of each repeated letter and obtain

11! Cenga

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11! Cenga

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4!4!2!Cenga

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4!4!2!�Cen

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� 34,650Cenga

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34,650

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FINDING THE NUMBER OF DISTINGUISHABLE PERMUTATIONSLearn

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FINDING THE NUMBER OF DISTINGUISHABLE PERMUTATIONSFind the number of distinguishable permutations of the letters in the word

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Find the number of distinguishable permutations of the letters in the wordLe

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DISTINGUISHABLE PERMUTATIONS OF IDENTICAL ITEMS

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ing DISTINGUISHABLE PERMUTATIONS OF IDENTICAL ITEMS

(or arrangements) of

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(or arrangements) of items are identical, and

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items are identical, andThat is, to find the number of distinguishable permutations,

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ing

That is, to find the number of distinguishable permutations,divide the total factorial by the factorial of each repeated item.

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divide the total factorial by the factorial of each repeated item.

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ing

24. Three hundred people buy raffle tickets. Three winningtickets will be drawn at random.

a. If first prize is $100, second prize is $50, and thirdprize is $20, in how many different ways can theprizes be awarded?

b. If each prize is $50, in how many different wayscan the prizes be awarded?

25. A group of nine women and six men must select a four-person committee. How many committees are possibleif it must consist of the following?

a. two women and two men

b. any mixture of men and women

c. a majority of women

26. A group of ten seniors, eight juniors, six sophomores,and five freshmen must select a committee of four.How many committees are possible if the committeemust contain the following:

a. one person from each class

b. any mixture of the classes

c. exactly two seniors

Exercises 27–32 refer to a deck of fifty-two playing cards (jokersnot allowed). If you are unfamiliar with playing cards, see the endof Section 3.1 for a description of a standard deck.

27. How many five-card poker hands are possible?

28. a. How many five-card poker hands consisting of allhearts are possible?

b. How many five-card poker hands consisting of allcards of the same suit are possible?

b. If their names are put in alphabetical order todetermine who goes first, second, and so on, howmany presentation orders are possible?

18. An English class consists of twenty-three students, andthree are to be chosen to give speeches in a school com-petition. In how many different ways can the teacherchoose the team, given the following conditions?

a. The order of the speakers is important.

b. The order of the speakers is not important.

19. In how many ways can the letters in the word “school”be arranged? (See the photograph below.)

20. A committee of four is to be selected from a group ofsixteen people. How many different committees arepossible, given the following conditions?

a. There is no distinction between the responsibilitiesof the members.

b. One person is the chair, and the rest are generalmembers.

c. One person is the chair, one person is the secretary,one person is responsible for refreshments, and oneperson cleans up after meetings.

21. A softball league has thirteen teams. If every team mustplay every other team once in the first round of leagueplay, how many games must be scheduled?

22. In a group of eighteen people, each person shakeshands once with each other person in the group. Howmany handshakes will occur?

23. A softball league has thirteen teams. How many differ-ent end-of-the-season rankings of first, second, andthird place are possible (disregarding ties)?

2.4 Exercises 115

Exercise 19: Right letters, wrong order. SHCOOL is painted along the newly paved roadleading to Southern Guilford High School on Drake Road Monday, August 9, 2010, inGreensboro, N.C.

AP P

hoto

/New

s &

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ord,

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Exercises 27–32 refer to a deck of fifty-two playing cards (jokers

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Exercises 27–32 refer to a deck of fifty-two playing cards (jokersnot allowed). If you are unfamiliar with playing cards, see the end

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not allowed). If you are unfamiliar with playing cards, see the endof Section 3.1 for a description of a standard deck.

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of Section 3.1 for a description of a standard deck.

27.

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27.In a group of eighteen people, each person shakes

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In a group of eighteen people, each person shakeshands once with each other person in the group. How

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hands once with each other person in the group. How

teams. How many differ-Cenga

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teams. How many differ-ent end-of-the-season rankings of first, second, andCen

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ent end-of-the-season rankings of first, second, and

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ing any mixture of men and women

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ing any mixture of men and women

a majority of women

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ing a majority of women

A group of ten seniors, eight juniors, six sophomores,

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ing

A group of ten seniors, eight juniors, six sophomores,and five freshmen must select a committee of four.

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ing

and five freshmen must select a committee of four.How many committees are possible if the committee

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ing

How many committees are possible if the committeemust contain the following:

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ing

must contain the following:

one person from each class

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ing

one person from each class

any mixture of the classesLearn

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any mixture of the classes

exactly two seniorsLearn

ing

exactly two seniors

g. Use the pattern described in part (f) to predict the sumof the entries in the sixth row of Pascal’s triangle.

h. Find the sum of the entries in the sixth row of Pascal’sTriangle. Was your prediction in part (g) correct?

i. Find the sum of the entries in the nth row of Pascal’sTriangle.

40. a. Add adjacent entries of the sixth row of Pascal’sTriangle to obtain the seventh row.

b. Find 6Cr for r � 0, 1, 2, 3, 4, 5, and 6.

c. How are the answers to parts (a) and (b) related?

41. Use Pascal’s Triangle to answer the following.

a. In which row would you find the value of 4C2?

b. In which row would you find the value of nCr?

c. Is 4C2 the second number in the fourth row?

d. Is 4C2 the third number in the fifth row?

e. What is the location of nCr? Why?

42. Given the set S � {a, b, c, d}, answer the following.

a. How many one-element subsets does S have?

b. How many two-element subsets does S have?

c. How many three-element subsets does S have?

d. How many four-element subsets does S have?

e. How many zero-element subsets does S have?

f. How many subsets does S have?

g. If n(S) � k, how many subsets will S have?

In Exercises 43–50, find the number of permutations of the lettersin each word.

43. ALASKA 44. ALABAMA

45. ILLINOIS 46. HAWAII

47. INDIANA 48. TENNESSEE

49. TALLAHASSEE 50. PHILADELPHIA

The words in each of Exercises 51–54 are homonyms (words thatare pronounced the same but have different meanings). Find thenumber of permutations of the letters in each word.

51. a. PIER b. PEER

52. a. HEAR b. HERE

53. a. STEAL b. STEEL

54. a. SHEAR b. SHEER



55. Suppose you want to know how many ways r itemscan be selected from a group of n items. Whatdetermines whether you should calculate nPr or nCr?

56. For any given values of n and r, which is larger, nPr or

nCr? Why?

29. a. How many five-card poker hands containingexactly three aces are possible?

b. How many five-card poker hands containing threeof a kind are possible?

30. a. How many five-card poker hands consisting ofthree kings and two queens are possible?

b. How many five-card poker hands consisting of threeof a kind and a pair (a full house) are possible?

31. How many five-card poker hands containing two pairare possible?

HINT: You must select two of the thirteen ranks, thenselect a pair of each, and then one of the remainingcards.

32. How many five-card poker hands containing exactlyone pair are possible?

HINT: After selecting a pair, you must select three ofthe remaining twelve ranks and then select one card ofeach.

33. A 6�53 lottery requires choosing six of the numbers 1through 53. How many different lottery tickets can youchoose? (Order is not important, and the numbers donot repeat.)

34. A 7�39 lottery requires choosing seven of the numbers1 through 39. How many different lottery tickets canyou choose? (Order is not important, and the numbersdo not repeat.)

35. A 5�36 lottery requires choosing five of the numbers 1through 36. How many different lottery tickets can youchoose? (Order is not important, and the numbers donot repeat.)

36. A 6�49 lottery requires choosing six of the numbers 1through 49. How many different lottery tickets can youchoose? (Order is not important, and the numbers donot repeat.)

37. Which lottery would be easier to win, a 6�53 or a5�36? Why?

HINT: See Exercises 33 and 35.

38. Which lottery would be easier to win, a 6�49 or a7�39? Why?

HINT: See Exercises 34 and 36.39. a. Find the sum of the entries in the first row of

Pascal’s Triangle.

b. Find the sum of the entries in the second row ofPascal’s Triangle.

c. Find the sum of the entries in the third row ofPascal’s Triangle.

d. Find the sum of the entries in the fourth row ofPascal’s Triangle.

e. Find the sum of the entries in the fifth row ofPascal’s Triangle.

f. Is there a pattern to the answers to parts (a)–(e)? Ifso, describe the pattern you see.


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y 49 lottery requires choosing six of the numbers 1

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y 49 lottery requires choosing six of the numbers 1through 49. How many different lottery tickets can you

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y through 49. How many different lottery tickets can youchoose? (Order is not important, and the numbers do

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y choose? (Order is not important, and the numbers do

Which lottery would be easier to win, a 6

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y Which lottery would be easier to win, a 6

See Exercises 33 and 35.

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See Exercises 33 and 35.

Which lottery would be easier to win, a 6Propert

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Which lottery would be easier to win, a 6

of through 36. How many different lottery tickets can you

of through 36. How many different lottery tickets can youchoose? (Order is not important, and the numbers do

of choose? (Order is not important, and the numbers do

49 lottery requires choosing six of the numbers 1of 49 lottery requires choosing six of the numbers 1

Cenga

ge How many two-element subsets does

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ge How many two-element subsets does

How many three-element subsets does

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ge How many three-element subsets does

How many four-element subsets does

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How many four-element subsets does

e.

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e. How many zero-element subsets does

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ge

How many zero-element subsets does

f.

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f. How many subsets does

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How many subsets does

g.

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ge

g. If

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If n

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n

In Exercises 43–50, find the number of permutations of the letters

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In Exercises 43–50, find the number of permutations of the lettersin each word.Cen

gage

in each word.36 lottery requires choosing five of the numbers 1

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36 lottery requires choosing five of the numbers 1through 36. How many different lottery tickets can youCen

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through 36. How many different lottery tickets can youchoose? (Order is not important, and the numbers doCen

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choose? (Order is not important, and the numbers do

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ing In which row would you find the value of

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In which row would you find the value of

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the second number in the fourth row?

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ing

the second number in the fourth row?

the third number in the fifth row?

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ing

the third number in the fifth row?

What is the location of

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ing

What is the location of n

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ing

nC

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ing

Cr

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ing

r? Why?

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ing

? Why?

�

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ing

� {a, b, c, d}, answer the following.

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ing

{a, b, c, d}, answer the following.

How many one-element subsets does

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ing


How many two-element subsets does Learn

ing

How many two-element subsets does

How many three-element subsets does Learn

ing


d. A and D; B and C; E and F

e. A and D; B and E; C and F

59. Which of the following teams must team B playsecond?

a. A b. C c. D d. E e. F

60. The last set of games could be between which teams?

a. A and B; C and F; D and E

b. A and C; B and F; D and E

c. A and D; B and C; E and F

d. A and E; B and C; D and F

e. A and F; B and E; C and D

61. If team D wins five games, which of the followingmust be true?

a. Team A loses five games.

b. Team A wins four games.

c. Team A wins its first game.

d. Team B wins five games.

e. Team B loses at least one game.

Web Project

62. Write a research paper on any historical topic referredto in this section or in a previous section. Following isa partial list of topics:● John Venn● Augustus De Morgan● Chu Shih-chieh

Some useful links for this web project are listed on thetext web site: www.cengage.com/math/johnson

Exercises 57–61 refer to the following: A baseball league has sixteams: A, B, C, D, E, and F. All games are played at 7:30 P.M. onFridays, and there are sufficient fields for each team to play agame every Friday night. Each team must play each otherteam exactly once, and the following conditions must be met:

● Team A plays team D first and team F second.● Team B plays team E first and team C third.● Team C plays team F first.

57. What is the total number of games that each team mustplay during the season?

a. 3 b. 4 c. 5 d. 6 e. 7

58. On the first Friday, which of the following pairs ofteams play each other?

a. A and B; C and F; D and E

b. A and B; C and E; D and F

c. A and C; B and E; D and F

2.5 Infinite Sets 117

2.5 Infinite Sets

Objectives

• Determine whether two sets are equivalent

• Establish a one-to-one correspondence between the elements of two sets

• Determine the cardinality of various infinite sets.

WARNING: Many leading nineteenth-century mathematicians and philosophersclaim that the study of infinite sets may be dangerous to your mental health.

Consider the sets E and N, where E � {2, 4, 6, . . .} and N � {1, 2, 3, . . .}. Bothare examples of infinite sets (they “go on forever”). E is the set of all even count-ing numbers, and N is the set of all counting (or natural) numbers. Because every


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element of E is an element of N, E is a subset of N. In addition, N contains ele-ments not in E; therefore, E is a proper subset of N. Which set is “bigger,” E orN? Intuition might lead many people to think that N is twice as big as E becauseN contains all the even counting numbers and all the odd counting numbers. Notso! According to the work of Georg Cantor (considered by many to be the fatherof set theory), N and E have exactly the same number of elements! This seemingparadox, a proper subset that has the same number of elements as the set fromwhich it came, caused a philosophic uproar in the late nineteenth century. (Hence,the warning at the beginning of this section.) To study Cantor’s work (which isnow accepted and considered a cornerstone in modern mathematics), we mustfirst investigate the meaning of a one-to-one correspondence and equivalent sets.

One-to-One Correspondence

Is there any relationship between the sets A � {one, two, three} and B � {Pontiac,Chevrolet, Ford}? Although the sets contain different types of things (numbersversus automobiles), each contains the same number of things; they are the samesize. This relationship (being the same size) forms the basis of a one-to-one corre-spondence. A one-to-one correspondence between the sets A and B is a pairing upof the elements of A and B such that each element of A is paired up with exactlyone element of B, and vice versa, with no element left out. For instance, the ele-ments of A and B might be paired up as follows:

one two three

D D D

Pontiac Chevrolet Ford

(Other correspondences, or matchups, are possible.) If two sets have the same car-dinal number, their elements can be put into a one-to-one correspondence. When-ever a one-to-one correspondence exists between the elements of two sets A and B,the sets are equivalent (denoted by A � B). Hence, equivalent sets have the samenumber of elements.

If two sets have different cardinal numbers, it is not possible to construct aone-to-one correspondence between their elements. The sets C � {one, two} andB � {Pontiac, Chevrolet, Ford} do not have a one-to-one correspondence; no mat-ter how their elements are paired up, one element of B will always be left over(B has more elements; it is “bigger”):

one two

D D


The sets C and B are not equivalent.Given two sets A and B, if any one of the following statements is true, then

the other statements are also true:

1. There exists a one-to-one correspondence between the elements of A and B.2. A and B are equivalent sets.3. A and B have the same cardinal number; that is, n(A) � n(B).

EXAMPLE 1 DETERMINING WHETHER TWO SETS ARE EQUIVALENT Determinewhether the sets in each of the following pairs are equivalent. If they areequivalent, list a one-to-one correspondence between their elements.

a. A � {John, Paul, George, Ringo};B � {Lennon, McCartney, Harrison, Starr}

b. C � {a, b, x, d}; D � {I, O, �}c. A � {1, 2, 3, . . . , 48, 49, 50}; B � {1, 3, 5, . . . , 95, 97, 99}

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SOLUTION a. If sets have the same cardinal number, they are equivalent. Now, n(A) � 4 and n(B) � 4;therefore, A � B.

Because A and B are equivalent, their elements can be put into a one-to-onecorrespondence. One such correspondence follows:

John Paul George Ringo

D D D D

Lennon McCartney Harrison Starr

b. Because n(C) � 4 and n(D) � 3, C and D are not equivalent.c. A consists of all natural numbers from 1 to 50, inclusive. Hence, n(A) � 50. B consists

of all odd natural numbers from 1 to 99, inclusive. Since half of the natural numbers

Georg Ferdinand LudwigPhilip Cantor was born in

St. Petersburg, Russia. His fatherwas a stockbroker and wantedhis son to become an engineer;his mother was an artist and mu-sician. Several of Cantor’s mater-nal relatives were accomplishedmusicians; in his later years,Cantor often wondered how his lifewould have turned out if he had becomea violinist instead of pursuing a controver-sial career in mathematics.

Following his father’s wishes,Cantor began his engineering studiesat the University of Zurich in 1862.However, after one semester, he de-cided to study philosophy and puremathematics. He transferred to theprestigious University of Berlin, studiedunder the famed mathematicians KarlWeierstrass, Ernst Kummer, andLeopold Kronecker, and received hisdoctorate in 1867. Two years later,Cantor accepted a teaching positionat the University of Halle and remainedthere until he retired in 1913.

Cantor’s treatises on set theory andthe nature of infinite sets were first pub-lished in 1874 in Crelle’s Journal, whichwas influential in mathematical circles.On their publication, Cantor’s theoriesgenerated much controversy amongmathematicians and philosophers. Para-doxes concerning the cardinal numbers

of infinite sets, the na-ture of infinity, andCantor’s form of logicwere unsettling tomany, including Can-tor’s former teacherLeopold Kronecker. Infact, some felt thatCantor’s work was notjust revolutionary butactually dangerous.Kronecker led the

attack on Cantor’s theories. He wasan editor of Crelle’s Journal and heldup the publication of one of Cantor’ssubsequent articles for so long thatCantor refused to publish ever againin the Journal. In addition, Kroneckerblocked Cantor’s efforts to obtain ateaching position at the University ofBerlin. Even though Cantor was at-tacked by Kronecker and hisfollowers, others respected him. Real-izing the importance of communica-tion among scholars, Cantorfounded the Association of GermanMathematicians in 1890 and servedas its president for many years. In ad-dition, Cantor was instrumental inorganizing the first InternationalCongress of Mathematicians, held inZurich in 1897.

As a result of the repeated at-tacks on him and his work, Cantorsuffered many nervous breakdowns,the first when he was thirty-nine. Hedied in a mental hospital in Halleat the age of seventy-three, never

HistoricalNote

GEORG CANTOR, 1845–1918

Written in 1874, Cantor’s first major paper on thetheory of sets, Über eine Eigenshaft des Inbegriffesaller reellen algebraischen Zahlen (On a Property ofthe System of All the Real Algebraic Numbers),sparked a major controversy concerning the natureof infinite sets. To gather international support for histheory, Cantor had his papers translated into French.This 1883 French version of Cantor’s work waspublished in the newly formed journal ActaMathematica. Cantor’s works were translated intoEnglish during the early twentieth century.

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having received proper recognition forthe true value of his discoveries. Mod-ern mathematicians believe that Can-tor’s form of logic and his concepts ofinfinity revolutionized all of mathemat-ics, and his work is now considered acornerstone in its development.

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from 1 to 100 are odd (and half are even), there are fifty (100 � 2 � 50) odd naturalnumbers less than 100; that is, n(B) � 50. Because A and B have the same cardinal num-ber, A � B.

Many different one-to-one correspondences may be established between the ele-ments of A and B. One such correspondence follows:

A � {1, 2, 3, . . . , n, . . . , 48, 49, 50}

D D D . . . D . . . D D D

B � {1, 3, 5, . . . , (2n � 1), . . . , 95, 97, 99}

That is, each natural number n � A is paired up with the odd number (2n � 1) � B.The n 4 (2n � 1) part is crucial because it shows each individual correspondence.For example, it shows that 13 � A corresponds to 25 � B (n � 13, so 2n � 26 and2n � 1 � 25). Likewise, 69 � B corresponds to 35 � A (2n � 1 � 69, so 2n � 70 andn � 35).

As we have seen, if two sets have the same cardinal number, they are equiv-alent, and their elements can be put into a one-to-one correspondence. Conversely,if the elements of two sets can be put into a one-to-one correspondence, the setshave the same cardinal number and are equivalent. Intuitively, this result appearsto be quite obvious. However, when Georg Cantor applied this relationship to infi-nite sets, he sparked one of the greatest philosophical debates of the nineteenthcentury.

Countable Sets

Consider the set of all counting numbers N � {1, 2, 3, . . .}, which consists of aninfinite number of elements. Each of these numbers is either odd or even. DefiningO and E as O � {1, 3, 5, . . .} and E � {2, 4, 6, . . .}, we have O ¨ E � � andO ´ E � N; the sets O and E are mutually exclusive, and their union forms the en-tire set of all counting numbers. Obviously, N contains elements that E does not. Aswe mentioned earlier, the fact that E is a proper subset of N might lead people tothink that N is “bigger” than E. In fact, N and E are the “same size”; N and E eachcontain the same number of elements.

Recall that two sets are equivalent and have the same cardinal number if theelements of the sets can be matched up via a one-to-one correspondence. To showthe existence of a one-to-one correspondence between the elements of two setsof numbers, we must find an explicit correspondence between the general ele-ments of the two sets. In Example 1(c), we expressed the general correspondenceas n 4 (2n � 1).

EXAMPLE 2 FINDING A ONE-TO-ONE CORRESPONDENCE BETWEEN TWOINFINITE SETS

a. Show that E � {2, 4, 6, 8, . . .} and N � {1, 2, 3, 4, . . .} are equivalent sets.b. Find the element of N that corresponds to 1430 � E.c. Find the element of N that corresponds to x � E.

SOLUTION a. To show that E � N, we must show that there exists a one-to-one correspondence betweenthe elements of E and N. The elements of E and N can be paired up as follows:

N � {1, 2, 3, 4, . . . , n, . . .}

D D D D . . . D

E � {2, 4, 6, 8, . . . , 2n, . . .}

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of infinite number of elements. Each of these numbers is either odd or even. Defining

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alent, and their elements can be put into a one-to-one correspondence. Conversely,if the elements of two sets can be put into a one-to-one correspondence, the sets

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Any natural number n � N corresponds with the even natural number 2n � E. Becausethere exists a one-to-one correspondence between the elements of E and N, the sets Eand N are equivalent; that is, E � N.

b. 1430 � 2n � E, so n � � 715 � N. Therefore, 715 � N corresponds to 1430 � E.c. x � 2n � E, so n � � N. Therefore, n � � N corresponds to x � 2n � E.

We have just seen that the set of even natural numbers is equivalent to the setof all natural numbers. This equivalence implies that the two sets have the samenumber of elements! Although E is a proper subset of N, both sets have the samecardinal number; that is, n(E) � n(N ). Settling the controversy sparked by thisseeming paradox, mathematicians today define a set to be an infinite set if it canbe placed in a one-to-one correspondence with a proper subset of itself.

How many counting numbers are there? How many even counting numbersare there? We know that each set contains an infinite number of elements andthat n(N) � n(E), but how many is that? In the late nineteenth century, Georg Cantordefined the cardinal number of the set of counting numbers to be ℵ0 (read “aleph-null”). Cantor utilized Hebrew letters, of which aleph, ℵ, is the first. Consequently,the proper response to “How many counting numbers are there?” is “There are aleph-null of them”; n(N) � ℵ0. Any set that is equivalent to the set of counting numbershas cardinal number ℵ0. A set is countable if it is finite or if it has cardinality ℵ0.

Cantor was not the first to ponder the paradoxes of infinite sets. Hundredsof years before, Galileo had observed that part of an infinite set contained asmany elements as the whole set. In his monumental Dialogue Concerning theTwo Chief World Systems (1632), Galileo made a prophetic observation:“There are as many (perfect) squares as there are (natural) numbers becausethey are just as numerous as their roots.” In other words, the elements of the setsN � {1, 2, 3, . . . , n, . . .} and S � {12, 22, 32, . . . , n2, . . .} can be put into a one-to-one correspondence (n 4 n2). Galileo pondered which of the sets (perfectsquares or natural numbers) was “larger” but abandoned the subject because hecould find no practical application of this puzzle.

EXAMPLE 3 SHOWING THAT THE SET OF INTEGERS IS COUNTABLE Considerthe following one-to-one correspondence between the set I of all integers and theset N of all natural numbers:

N � {1, 2, 3, 4, 5, . . .}D D D D D

I � {0, 1, �1, 2, �2, . . .}

where an odd natural number n corresponds to a nonpositive integer and aneven natural number n corresponds to a positive integer .

a. Find the 613th integer; that is, find the element of I that corresponds to 613 � N.b. Find the element of N that corresponds to 853 � I.c. Find the element of N that corresponds to �397 � I.d. Find n(I), the cardinal number of the set I of all integers.e. Is the set of integers countable?

SOLUTION a. 613 � N is odd, so it corresponds to If you continued countingthe integers as shown in the above correspondence, �306 would be the 613th integer inyour count.

b. 853 � I is positive, so

multiplying by 2

1,706 � N corresponds to 853 � I.This means that 853 is the 1,706th integer.

n � 1,706

853 �n

2

1 � 6132 � �612

2 � �306.

n2

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x2

x2

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c. �397 � I is negative, so

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795 � N corresponds to �397 � I.This means that �397 is the 795th integer.

d. The given one-to-one correspondence shows that I and N have the same (infinite) num-ber of elements; n(I) � n(N ). Because n(N ) � ℵ0, the cardinal number of the set of allintegers is n(I) � ℵ0.

e. By definition, a set is called countable if it is finite or if it has cardinality ℵ0. The set ofintegers has cardinality ℵ0, so it is countable. This means that we can “count off” all ofthe integers, as we did in parts (a), (b), and (c).

We have seen that the sets N (all counting numbers), E (all even countingnumbers), and I (all integers) contain the same number of elements, ℵ0. Whatabout a set containing fractions?

EXAMPLE 4 DETERMINING WHETHER THE SET OF POSITIVE RATIONAL NUM-BERS IS COUNTABLE Determine whether the set P of all positive rationalnumbers is countable.

SOLUTION The elements of P can be systematically listed in a table of rows and columns asfollows: All positive rational numbers whose denominator is 1 are listed in the firstrow, all positive rational numbers whose denominator is 2 are listed in the secondrow, and so on, as shown in Figure 2.43.

Each positive rational number will appear somewhere in the table. For in-stance, will be in row 66 and column 125. Note that not all the entries in Fig-ure 2.43 are in lowest terms; for instance, , , , and so on are all equal to .1

248

36

24

12566

n � 795

�795 � �n

�794 � 1 � n

�397 �1 � n

2


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12

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32

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13

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13

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14

24

34

44

15

25

35

45

52

Start: 11 2

131

41

51

53

54

55

A list of all positive rationalnumbers.

FIGURE 2.43 The circled rational numbers arenot in lowest terms; they areomitted from the list.

FIGURE 2.44

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(all integers) contain the same number of elements,


Consequently, to avoid listing the same number more than once, an entry that isnot in lowest terms must be eliminated from our list. To establish a one-to-one corre-spondence between P and N, we can create a zigzag diagonal pattern as shown by thearrows in Figure 2.44. Starting with , we follow the arrows and omit any number thatis not in lowest terms (the circled numbers in Figure 2.44). In this manner, a list of allpositive rational numbers with no repetitions is created. Listing the elements of P inthis order, we can put them in a one-to-one correspondence with N:

N � {1, 2, 3, 4, 5, 6, 7, 8, 9, 10,11, . . .}D D D D D D D DD D D

Any natural number n is paired up with the positive rational number found bycounting through the “list” given in Figure 2.44. Conversely, any positive rationalnumber is located somewhere in the list and is paired up with the counting numbercorresponding to its place in the list.

Therefore, P � N, so the set of all positive rational numbers is countable.

Uncountable Sets

Every infinite set that we have examined so far is countable; each can be put into aone-to-one correspondence with the set of all counting numbers and consequentlyhas cardinality ℵ0. Do not be misled into thinking that all infinite sets are counta-ble! By utilizing a “proof by contradiction,” Georg Cantor showed that the infiniteset A � {x � 0 � x � 1} is not countable. This proof involves logic that is differentfrom what you are used to. Do not let that intimidate you.

Assume that the set A � {x � 0 � x � 1} is countable; that is, assume that n(A) � ℵ0. This assumption implies that the elements of A and N can be put into aone-to-one correspondence; each a � A can be listed and counted. Because the ele-ments of A are nonnegative real numbers less than 1, each an � 0. . . . .Say, for instance, the numbers in our list are

a1 � 0.3750000 . . . the first element of Aa2 � 0.7071067 . . . the second element of Aa3 � 0.5000000 . . . the third element of Aa4 � 0.6666666 . . . and so on.

The assumption that A is countable implies that every element of A appearssomewhere in the above list. However, we can create an element of A (call it b) that isnot in the list. We build b according to the “diagonal digits” of the numbers in our listand the following rule: If the digit “on the diagonal” is not zero, put a 0 in thecorresponding place in b; if the digit “on the diagonal” is zero, put a 1 in the corre-sponding place in b.

The “diagonal digits” of the numbers in our list are as follows:

a1 � 0. 750000 . . .

a2 � 0.7 71067 . . .

a3 � 0.50 0000 . . .

a4 � 0.666 666 . . .

Because the first digit on the diagonal is 3, the first digit of b is 0. Because thesecond digit on the diagonal is 0, the second digit of b is 1. Using all the “diagonal

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so the set of all positive rational numbers is countable.

digits” of the numbers in our list, we obtain b � 0.0110. . . . Because 0 � b � 1, itfollows that b � A. However, the number b is not on our list of all elements of A.This is because

b � a1 (b and a1 differ in the first decimal place)b � a2 (b and a2 differ in the second decimal place)b � a3 (b and a3 differ in the third decimal place), and so on

This contradicts the assumption that the elements of A and N can be put intoa one-to-one correspondence. Since the assumption leads to a contradiction, theassumption must be false; A � {x � 0 � x � 1} is not countable. Therefore,n(A) � n(N ). That is, A is an infinite set and n(A) � ℵ0.

An infinite set that cannot be put into a one-to-one correspondence with N issaid to be uncountable. Consequently, an uncountable set has more elements thanthe set of all counting numbers. This implies that there are different magnitudes ofinfinity! To distinguish the magnitude of A from that of N, Cantor denoted the car-dinality of A � {x � 0 � x � 1} as n(A) � c (c for continuum). Thus, Cantorshowed that ℵ0 � c. Cantor went on to show that A was equivalent to the entire setof all real numbers, that is, A � t. Therefore, n(t) � c.

Although he could not prove it, Cantor hypothesized that no set could have acardinality between ℵ0 and c. This famous unsolved problem, labeled the Contin-uum Hypothesis, baffled mathematicians throughout the first half of the twentiethcentury. It is said that Cantor suffered a devastating nervous breakdown in 1884when he announced that he had a proof of the Continuum Hypothesis only to de-clare the next day that he could show the Continuum Hypothesis to be false!

The problem was finally “solved” in 1963. Paul J. Cohen demonstrated thatthe Continuum Hypothesis is independent of the entire framework of set theory;that is, it can be neither proved nor disproved by using the theorems of set theory.Thus, the Continuum Hypothesis is not provable.

Although no one has produced a set with cardinality between ℵ0 and c, manysets with cardinality greater than c have been constructed. In fact, modern mathe-maticians have shown that there are infinitely many magnitudes of infinity! Usingsubscripts, these magnitudes, or cardinalities, are represented by ℵ0, ℵ1, ℵ2, . . .and have the property that ℵ0 � ℵ1 � ℵ2 � . . . . In this sense, the set N of all nat-ural numbers forms the “smallest” infinite set. Using this subscripted notation, theContinuum Hypothesis implies that c � ℵ1; that is, given that N forms the small-est infinite set, the set t of all real numbers forms the next “larger” infinite set.

Points on a Line

When students are first exposed to the concept of the real number system, a num-ber line like the one in Figure 2.45 is inevitably introduced. The real number sys-tem, denoted by t, can be put into a one-to-one correspondence with all points ona line, such that every real number corresponds to exactly one point on a line andevery point on a line corresponds to exactly one real number. Consequently, any(infinite) line contains c points. What about a line segment? For example, howmany points does the segment [0, 1] contain? Does the segment [0, 2] containtwice as many points as the segment [0, 1]? Once again, intuition can lead to erro-neous conclusions when people are dealing with infinite sets.

EXAMPLE 5 SHOWING THAT LINE SEGMENTS OF DIFFERENT LENGTHS AREEQUIVALENT SETS OF POINTS Show that the line segments [0, 1] and [0, 2] are equivalent sets of points.

SOLUTION Because the segment [0, 2] is twice as long as the segment [0, 1], intuition mighttell us that it contains twice as many points. Not so! Recall that two sets are


–2 –1 0 2�

1

The real number line.

FIGURE 2.45

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equivalent (and have the same cardinal number) if their elements can be put into aone-to-one correspondence.

On a number line, let A represent the point 0, let B represent 1, and let Crepresent 2, as shown in Figure 2.46. Our goal is to develop a one-to-one corre-spondence between the elements of the segments AB and AC. Now draw thesegments separately, with AB above AC, as shown in Figure 2.47. (To distin-guish the segments from each other, point A of segment AB has been relabeledas point A.)

Extend segments AA and CB so that they meet at point D, as shown inFigure 2.48. Any point E on AB can be paired up with the unique point F on ACformed by the intersection of lines DE and AC, as shown in Figure 2.49. Con-versely, any point F on segment AC can be paired up with the unique point E onAB formed by the intersection of lines DF and AB. Therefore, a one-to-onecorrespondence exists between the two segments, so [0, 1] � [0, 2]. Conse-quently, the interval [0, 1] contains exactly the same number of points as theinterval [0, 2]!

A B C

0 1 2

The interval [0, 2].

FIGURE 2.46

A C

A B

The intervals [A, B] and [A, C ].

FIGURE 2.47

D

A

A B

C

D

E

FA

A B

C

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FIGURE 2.48 A one-to-one correspondencebetween line segments.

FIGURE 2.49

Even though the segment [0, 2] is twice as long as the segment [0, 1], eachsegment contains exactly the same number of points. The method used in Exam-ple 5 can be applied to any two line segments. Consequently, all line segments,regardless of their length, contain exactly the same number of points; a line seg-ment 1 inch long has exactly the same number of points as a segment 1 milelong! Once again, it is easy to see why Cantor’s work on the magnitude of in-finity was so unsettling to many scholars.

Having concluded that all line segments contain the same number ofpoints, we might ask how many points that is. What is the cardinal number?Given any line segment AB, it can be shown that n(AB) � c; the points of a linesegment can be put into a one-to-one correspondence with the points of a line.Consequently, the interval [0, 1] contains the same number of elements as theentire real number system.

If things seem rather strange at this point, keep in mind that Cantor’s pio-neering work produced results that puzzled even Cantor himself. In a paperwritten in 1877, Cantor constructed a one-to-one correspondence between thepoints in a square (a two-dimensional figure) and the points on a line segment(a one-dimensional figure). Extending this concept, he concluded that a linesegment and the entire two-dimensional plane contain exactly the same numberof points, c. Communicating with his colleague Richard Dedekind, Cantorwrote, “I see it, but I do not believe it.” Subsequent investigation has shownthat the number of points contained in the interval [0, 1] is the same as the num-ber of points contained in all of three-dimensional space! Needless to say,Cantor’s work on the cardinality of infinity revolutionized the world of modernmathematics.

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In Exercises 1–10, find the cardinal numbers of the sets in eachgiven pair to determine whether the sets are equivalent. If they areequivalent, list a one-to-one correspondence between their elements.

1. S � {Sacramento, Lansing, Richmond, Topeka}

C � {California, Michigan, Virginia, Kansas}

2. T � {Wyoming, Ohio, Texas, Illinois, Colorado}

P � {Cheyenne, Columbus, Austin, Springfield, Denver}

3. R � {a, b, c}; G � {a, b, x, d}

4. W � {I, II, III}; H � {one, two}

5. C � {3, 6, 9, 12, . . . , 63, 66}

D � {4, 8, 12, 16, . . . , 84, 88}

6. A � {2, 4, 6, 8, . . . , 108, 110}

B � {5, 10, 15, 20, . . . , 270, 275}

7. G � {2, 4, 6, 8, . . . , 498, 500}

H � {1, 3, 5, 7, . . . , 499, 501}

8. E � {2, 4, 6, 8, . . . , 498, 500}

F � {3, 6, 9, 12, . . . , 750, 753}

9. A � {1, 3, 5, . . . , 121, 123}

B � {125, 127, 129, . . . , 245, 247}

10. S � {4, 6, 8, . . . , 664, 666}

T � {5, 6, 7, . . . , 335, 336}

11. a. Show that the set O of all odd counting numbers, O � {1, 3, 5, 7, . . .}, and N � {1, 2, 3, 4, . . .} areequivalent sets.

b. Find the element of N that corresponds to 1,835 � O.

c. Find the element of N that corresponds to x � O.

d. Find the element of O that corresponds to 782 � N.

e. Find the element of O that corresponds to n � N.

12. a. Show that the set W of all whole numbers, W � {0, 1,2, 3, . . .}, and N � {1, 2, 3, 4, . . .} are equivalent sets.

b. Find the element of N that corresponds to 932 � W.

c. Find the element of N that corresponds to x � W.

d. Find the element of W that corresponds to 932 � N.

e. Find the element of W that corresponds to n � N.

13. a. Show that the set T of all multiples of 3, T � {3, 6, 9,12, . . .}, and N � {1, 2, 3, 4, . . .} are equivalent sets.

b. Find the element of N that corresponds to 936 � T.

c. Find the element of N that corresponds to x � T.

d. Find the element of T that corresponds to 936 � N.

e. Find the element of T that corresponds to n � N.

14. a. Show that the set F of all multiples of 5, F � {5, 10,15, 20, . . .}, and N � {1, 2, 3, 4, . . .} are equivalentsets.

b. Find the element of N that corresponds to 605 � F.

c. Find the element of N that corresponds to x � F.

d. Find the element of F that corresponds to 605 � N.

e. Find the element of F that corresponds to n � N.

15. Consider the following one-to-one correspondencebetween the set A of all even integers and the set N ofall natural numbers:

N � {1, 2, 3, 4, 5, . . .}

D D D D D

A � {0, 2, �2, 4, �4, . . .}

where an odd natural number n corresponds to thenonpositive integer 1 � n and an even natural numbern corresponds to the positive even integer n.

a. Find the 345th even integer; that is, find theelement of A that corresponds to 345 � N.

b. Find the element of N that corresponds to 248 � A.

c. Find the element of N that corresponds to �754 � A.

d. Find n(A).

16. Consider the following one-to-one correspondencebetween the set B of all odd integers and the set N of allnatural numbers:

N � {1, 2, 3, 4, 5, . . .}

D D D D D

B � {1, �1, 3, �3, 5, . . .}

where an even natural number n corresponds tothe negative odd integer 1 � n and an odd naturalnumber n corresponds to the odd integer n.

a. Find the 345th odd integer; that is, find the elementof B that corresponds to 345 � N.

b. Find the element of N that corresponds to 241 � B.

c. Find the element of N that corresponds to �759 � B.

d. Find n(B).

In Exercises 17–22, show that the given sets of points areequivalent by establishing a one-to-one correspondence.

17. the line segments [0, 1] and [0, 3]

18. the line segments [1, 2] and [0, 3]

19. the circle and square shown in Figure 2.50

HINT: Draw one figure inside the other.

FIGURE 2.50

2.5 Exercises

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and an even natural number

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and an even natural numbercorresponds to the positive even integer

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corresponds to the positive even integer

Find the 345th even integer; that is, find the

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Find the 345th even integer; that is, find thethat corresponds to 345

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Find the element of

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Chapter Review 127



24. What is the cardinal number of the “smallest” infiniteset? What set or sets have this cardinal number?

• history Questions

25. What aspect of Georg Cantor’s set theory caused controversy among mathematicians and philosophers?

26. What contributed to Cantor’s breakdown in 1884?

27. Who demonstrated that the Continuum Hypothesiscannot be proven? When?

Web Project

28. Write a research paper on any historical topic referredto in this section. Following is a partial list of topics:● Georg Cantor● Richard Dedekind● Paul J. Cohen● Bernhard Bolzano● Leopold Kronecker● the Continuum Hypothesis

Some useful links for this web project are listed on thetext web site: www.cengage.com/math/johnson

20. the rectangle and triangle shown in Figure 2.51


21. a circle of radius 1 cm and a circle of 5 cm


22. a square of side 1 cm and a square of side 5 cm


23. Show that the set of all real numbers between 0 and 1 has the same cardinality as the set of all realnumbers.

HINT: Draw a semicircle to represent the set of realnumbers between 0 and 1 and a line to represent theset of all real numbers, and use the method ofExample 5.

FIGURE 2.51

Chapter Review2TERMSaleph-nullcardinal numbercombinationcombinatoricscomplementcontinuumcountable set

De Morgan’s Lawsdistinguishable

permutationselementempty setequal setsequivalent setsfactorialFundamental Principle

of Counting

improper subsetinfinite setintersectionmutually exclusiveone-to-one

correspondencepermutationproper subsetroster notationset

set-builder notationset theorysubsettree diagramuncountable setunionuniversal setVenn diagramwell-defined set

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Chapter ReviewPropert

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of Cen

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Paul J. Cohen●

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● Bernhard Bolzano

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Who demonstrated that the Continuum Hypothesis

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cannot be proven? When?

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cannot be proven? When?

Web Project

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Write a research paper on any historical topic referred

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to in this section. Following is a partial list of topics:

Georg CantorLearn

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1. State whether the given set is well-defined.

a. the set of all multiples of 5.

b. the set of all difficult math problems

c. the set of all great movies

d. the set of all Oscar-winning movies

2. Given the sets

U � {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

A � {0, 2, 4, 6, 8}

B � {1, 3, 5, 7, 9}

find the following using the roster method.

a. A b. B

c. A ´ B d. A ¨ B

3. Given the sets A � {Maria, Nobuko, Leroy, Mickey,Kelly} and B � {Rachel, Leroy, Deanna, Mickey},find the following.

a. A ´ B b. A ¨ B

4. List all subsets of C � {Dallas, Chicago, Tampa}.Identify which subsets are proper and which areimproper.

5. Given n(U) � 61, n(A) � 32, n(B) � 26, and n(A ´ B) � 40, do the following.

a. Find n(A ¨ B).

b. Draw a Venn diagram illustrating the compositionof U.

In Exercises 6 and 7, use a Venn diagram like the one in Figure 2.15to shade in the region corresponding to the indicated set.

6. (A ´ B)

7. (A ¨ B)

In Exercises 8 and 9, use a Venn diagram like the one in Figure2.36 to shade in the region corresponding to the indicated set.

8. A ¨ (B ´ C)

9. (A ¨ B) ´ C

10. A survey of 1,000 college seniors yielded the followinginformation: 396 seniors favored capital punishment,531 favored stricter gun control, and 237 favored both.

a. How many favored capital punishment or strictergun control?

b. How many favored capital punishment but notstricter gun control?

c. How many favored stricter gun control but notcapital punishment?

d. How many favored neither capital punishment norstricter gun control?

11. A survey of recent college graduates yielded thefollowing information: 70 graduates earned a degreein mathematics, 115 earned a degree in education, 23earned degrees in both mathematics and education,

and 358 earned a degree in neither mathematics noreducation.a. What percent of the college graduates earned a

degree in mathematics or education?b. What percent of the college graduates earned a

degree in mathematics only?c. What percent of the college graduates earned a

degree in education only?12. A local anime fan club surveyed 67 of its members

regarding their viewing habits last weekend, and thefollowing information was obtained: 30 memberswatched an episode of Naruto, 44 watched an episode ofDeath Note, 23 watched an episode of Inuyasha,20 watched both Naruto and Inuyasha, 5 watchedNaruto and Inuyasha but not Death Note, 15 watchedboth Death Note and Inuyasha, and 23 watched onlyDeath Note.a. How many of the club members watched exactly

one of the shows?b. How many of the club members watched all three

shows?c. How many of the club members watched none of

the three shows?13. An exit poll yielded the following information

concerning people’s voting patterns on Propositions A,B, and C: 305 people voted yes on A, 95 voted yesonly on A, 393 voted yes on B, 192 voted yes only onB, 510 voted yes on A or B, 163 voted yes on C, 87voted yes on all three, and 249 voted no on all three.What percent of the voters voted yes on more than oneproposition?

14. Given the sets U � {a, b, c, d, e, f, g, h, i}, A �{b, d, f, g}, and B � {a, c, d, g, i}, use De Morgan’sLaws to find the following.a. (A ´ B) b. (A ¨ B)

15. Refer to the Venn diagram depicted in Figure 2.32.a. In a group of 100 Americans, how many have type

O or type A blood?b. In a group of 100 Americans, how many have type

O and type A blood?c. In a group of 100 Americans, how many have

neither type O nor type A blood?16. Refer to the Venn diagram depicted in Figure 2.28.

a. For a typical group of 100 Americans, fill in thecardinal number of each region in the diagram.

b. In a group of 100 Americans, how many have typeO blood or are Rh�?

c. In a group of 100 Americans, how many have typeO blood and are Rh�?

d. In a group of 100 Americans, how many haveneither type O blood nor are Rh�?

REVIEW EXERCISES


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y to shade in the region corresponding to the indicated set.

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information: 396 seniors favored capital punishment,

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b.

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shows?c.

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following information was obtained: 30 members

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following information was obtained: 30 membersNaruto

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Naruto, 44 watched an episode of

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, 44 watched an episode of, 23 watched an episode of

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Naruto and

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and Inuyasha

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InuyashaInuyasha

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Inuyasha but not

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but not Death Note

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Death NoteDeath Note

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Death Note and

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and Inuyasha

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one of the shows?

17. Sid and Nancy are planning their anniversary celebra-tion, which will include viewing an art exhibit, havingdinner, and going dancing. They will go either to theMuseum of Modern Art or to the New Photo Gallery;dine either at Stars, at Johnny’s, or at the Chelsea; and godancing either at Le Club or at Lizards.a. In how many different ways can Sid and Nancy

celebrate their anniversary?b. Construct a tree diagram to list all possible ways

in which Sid and Nancy can celebrate theiranniversary.

18. A certain model of pickup truck is available in fiveexterior colors, three interior colors, and three interiorstyles. In addition, the transmission can be eithermanual or automatic, and the truck can have either two-wheel or four-wheel drive. How many differentversions of the pickup truck can be ordered?

19. Each student at State University has a student I.D.number consisting of five digits (the first digit isnonzero, and digits can be repeated) followed by twoof the letters A, B, C, and D (letters cannot berepeated). How many different student numbers arepossible?

20. Find the value of each of the following.

a. (17 � 7)! b. (17 � 17)!

c. d.

21. In how many ways can you select three out of elevenitems under the following conditions?

a. Order of selection is not important.

b. Order of selection is important.

22. Find the value of each of the following.

a. 15P4 b. 15C4 c. 15P11

23. A group of ten women and twelve men must select athree-person committee. How many committees arepossible if it must consist of the following?

a. one woman and two men

b. any mixture of men and women

c. a majority of men

24. A volleyball league has ten teams. If every team mustplay every other team once in the first round of leagueplay, how many games must be scheduled?

25. A volleyball league has ten teams. How many differentend-of-the-season rankings of first, second, and thirdplace are possible (disregarding ties)?

26. Using a standard deck of fifty-two cards (no jokers),how many seven-card poker hands are possible?

27. Using a standard deck of fifty-two cards and two jokers,how many seven-card poker hands are possible?

28. A 6/42 lottery requires choosing six of the numbers 1through 42. How many different lottery tickets can youchoose?

27!

20!7!

82!

79!

Chapter Review 129

In Exercises 29 and 30, find the number of permutations of theletters in each word.

29. a. FLORIDA b. ARIZONA c. MONTANA

30. a. AFFECT b. EFFECT

31. What is the major difference between permutationsand combinations?

32. Use Pascal’s Triangle to answer the following.

a. In which entry in which row would you find thevalue of 7C3?

b. In which entry in which row would you find thevalue of 7C4?

c. How is the value of 7C3 related to the value of 7C4?Why?

d. What is the location of nCr? Why?

33. Given the set S � {a, b, c}, answer the following.

a. How many one-element subsets does S have?

b. How many two-element subsets does S have?

c. How many three-element subsets does S have?

d. How many zero-element subsets does S have?

e. How many subsets does S have?

f. How is the answer to part (e) related to n(S)?

In Exercises 34–36, find the cardinal numbers of the sets in eachgiven pair to determine whether the sets are equivalent. If they areequivalent, list a one-to-one correspondence between their elements.

34. A � {I, II, III, IV, V} and B � {one, two, three, four,five}

35. C � {3, 5, 7, . . . , 899, 901} and D � {2, 4, 6, . . . ,898, 900}

36. E � {Ronald} and F � {Reagan, McDonald}

37. a. Show that the set S of perfect squares, S � {1, 4, 9,16, . . . }, and N � {1, 2, 3, 4, . . . } are equivalentsets.

b. Find the element of N that corresponds to 841 � S.

c. Find the element of N that corresponds to x � S.

d. Find the element of S that corresponds to 144 � N.

e. Find the element of S that corresponds to n � N.

38. Consider the following one-to-one correspondencebetween the set A of all integer multiples of 3 and theset N of all natural numbers:

N � {1, 2, 3, 4, 5, . . .}

D D D D D

A � {0, 3, �3, 6, �6, . . .}

where an odd natural number n corresponds tothe nonpositive integer (1 � n), and an even naturalnumber n corresponds to the positive even integer n.

a. Find the element of A that corresponds to 396 � N.

b. Find the element of N that corresponds to 396 � A.

c. Find the element of N that corresponds to �153 � A.

d. Find n(A).

32

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given pair to determine whether the sets are equivalent. If they areequivalent, list a one-to-one correspondence between their elements.

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equivalent, list a one-to-one correspondence between their elements.

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39. Show that the line segments [0, 1] and [0, p] areequivalent sets of points by establishing a one-to-onecorrespondence.



40. What is the difference between proper and impropersubsets?

41. Explain the difference between {0} and �.

42. What is a factorial?

43. What is the difference between permutations andcombinations?


44. What roles did the following people play in thedevelopment of set theory and combinatorics?● Georg Cantor● Augustus De Morgan● Christian Kramp● Chu Shih-chieh● John Venn

Exercises 45–48 refer to the following: Two doctors in a localclinic are determining which days of the week they will be on call.Each day, Monday through Sunday, is to be assigned to one of twodoctors, A and B, such that the assignment is consistent with thefollowing conditions:

● No day is assigned to both doctors.● Neither doctor has more than four days.● Monday and Thursday must be assigned to the same doctor.● If Tuesday is assigned to doctor A, then so is Sunday.● If Saturday is assigned to doctor B, then Friday is not

assigned to doctor B.

45. Which one of the following could be a complete andaccurate list of the days assigned to doctor A?

a. Monday, Thursday

b. Monday, Tuesday, Sunday

c. Monday, Thursday, Sunday

d. Monday, Tuesday, Thursday

e. Monday, Thursday, Friday, Sunday

46. Which of the following cannot be true?

a. Thursday and Sunday are assigned to doctor A.

b. Friday and Saturday are assigned to doctor A.

c. Monday and Tuesday are assigned to doctor B.

d. Monday, Wednesday, and Sunday are assigned todoctor A.

e. Tuesday, Wednesday, and Saturday are assigned todoctor B.

47. If Friday and Sunday are both assigned to doctor B,how many different ways are there to assign the otherfive days to the doctors?

a. 1 b. 2 c. 3

d. 5 e. 6

48. If doctor A has four days, none of which is Monday,which of the following days must be assigned todoctor A?

a. Tuesday b. Wednesday c. Friday

d. Saturday e. Sunday


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d.

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doctor

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Monday, Thursday, Sunday

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Monday, Tuesday, Thursday

Monday, Thursday, Friday, Sunday

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Monday, Thursday, Friday, Sunday

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Which of the following cannot be true?

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Documents

Sets and Counting - Cengage › gpms › websites › 1133228607 › Ch 2... · 2011-09-20 · 70 CHAPTER 2 Sets and Counting Two sets are equalif they contain exactly the same elements.The