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FUNDAMENTAL PRINCIPLE OF COUNTING
If an operation can be performed in 'm' different ways and another operation in 'n' different ways then these two operations can be performed one after the other in 'mn' ways.
If an operation can be performed in 'm' different ways and another operation in 'n' different ways then either ofthese two operations can be performed in 'm+n' ways.(provided only one has to be done)
This principle can be extended to any number of operations
FACTORIAL 'n'
The continuous product of the first 'n' natural numbers is called factorial n and is deonoted by n! i.e, n!=1×2x3x…..x(n-1)xn.
PERMUTATION
An arrangement that can be formed by taking some or all of a finite set of things (or objects) is called a Permutation.
Order of the things is very important in case of permutation.
A permutation is said to be a Linear Permutation if the objects are arranged in a line. A linear permutation is simply called as a permutation.
A permutation is said to be a Circular Permutation if the objects are arranged in the form of a circle.
The number of (linear) permutations that can be formed by taking r things at a time from a set of n distinct things is denoted by .
.
NUMBER OF PERMUTATIONS UNDER CERTAIN CONDITIONS
1. Number of permutations of n different things, taken r at a time, when a particular thng is to be always included in each arrangement , is .
2. Number of permutations of n different things, taken r at a time, when a particular thing is never taken in each arrangement is .
3. Number of permutations of n different things, taken all at a time, when m specified things always come together is .
4. Number of permutations of n different things, taken all at a time, when m specified never come together is .
5. The number of permutations of n dissimilar things taken r at a time when k(< r) particular things always occur is .
6. The number of permutations of n dissimilar things taken r at a time when k particular things never occur is .
7. The number of permutations of n dissimilar things taken r at a time when repetition of things is allowed any number of times is .
8. The number of permutations of n different things, taken not more than r at a time, when each thing may occur any number of times is
.
9. The number of permutations of n different things taken not more than r at a time .
PERMUTATIONS OF SIMILAR THINGS
The number of permutations of n things taken all tat a time when p of
them are all alike and the rest are all different is .
If p things are alike of one type, q things are alike of other type, r things are alike of another type, then the number of permutations with p+q+r
things is .
CIRCULAR PERMUTATIONS
1. The number of circular permutations of n dissimilar things taken r at a
time is .
2. The number of circular permutations of n dissimilar things taken all at a time is .
3. The number of circular permutations of n things taken r at a time in one
direction is .
4. The number of circular permutations of n dissimilar things in clock-wise
direction = Number of permutations in anticlock-wise direction = .
COMBINATION
A selection that can be formed by taking some or all of a finite set of things( or objects) is called a Combination.
The number of combinations of n dissimilar things taken r at a time is
denoted by .
1.
2.
3.
4.
5. The number of combinations of n things taken r at a time in which
a)s particular things will always occur is .
b)s particular things will never occur is .
c)s particular things always occurs and p particular things never occur is .
DISTRIBUTION OF THINGS INTO GROUPS
1.Number of ways in which (m+n) items can be divided into two unequal
groups containing m and n items is .
2.The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is
not important is .
3.The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is
important is .
4.The number of ways in which (m+n+p) things can be divided into three
different groups of m,n, an p things respectively is
5.The required number of ways of dividing 3n things into three groups of n
each = .When the order of groups has importance then the required
number of ways= .
DIVISION OF IDENTICAL OBJECTS INTO GROUPS
The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is
The number of non-negative integral solutions of the equation .
The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is
The number of positive integral solutions of the equation .
The number of ways of choosing r objects from p objects of one kind, q objects of second kind, and so on is the coefficient of in the expansion
he number of ways of choosing r objects from p objects of one kind, q objects of second kind, and so on, such that one object of each kind may be included is the coefficient of is the coefficient of in the expansion
.
TOTAL NUMBER OF COMBINATIONS
1.The total number of combinations of things taken any number at a time when things are alike of one kind, things are alike of second kind…. things are alike of kind, is .
2.The total number of combinations of things taken one or more at a time when things are alike of one kind, things are alike of second kind…. things are alike of kind, is
.
SUM OF THE NUMBERS
Sum of the numbers formed by taking all the given n digits (excluding 0) is
Sum of the numbers formed by taking all the given n digits (including 0) is
Sum of all the r-digit numbers formed by taking the given n digits(excluding 0) is
Sum of all the r-digit numbers formed by taking the given n digits(including 0) is
DE-ARRANGEMENT:
The number of ways in which exactly r letters can be placed in wrongly addressed envelopes when n letters are placed in n addressed envelopes
is .
The number of ways in which n different letters can be placed in their n addressed envelopes so that al the letters are in the wrong envelopes is
.
IMPORTANT RESULTS TO REMEBER
In a plane if there are n points of which no three are collinear, then
1. The number of straight lines that can be formed by joining them is .
2. The number of triangles that can be formed by joining them is .
3. The number of polygons with k sides that can be formed by joining them is .
In a plane if there are n points out of which m points are collinear, then
1. The number of straight lines that can be formed by joining them is .
2. The number of triangles that can be formed by joining them is .
3. The number of polygons with k sides that can be formed by joining them is .
Number of rectangles of any size in a square of n x n is
Number of squares of any size in a square of n x n is
In a rectangle of p x q (p < q) number of rectangles of any size is
In a rectangle of p x q (p < q) number of squares of any size is
n straight lines are drawn in the plane such that no two lines are parallel and no three lines three lines are concurrent. Then the number of parts
into which these lines divide the plane is equal to .
Permutation and CombinationsTable of ContentsTutorial Examples Assignment
Tutorial
Permutations:
The ways in which a number of given objects can be arranged by taking
all of them or a specified number of objects out of them are called
PERMUTATIONS. Thus the number of permutations of three objects, viz.
a, b, and c, taking all of them at a time is 6 i.e., abc, acb, bcd, bac, cab
and cba.
The number of ways in which 2 objects can be taken and arranged out of
3 objects a, b and c is 6, viz. ab, ba, bc, cb, ac and ca. The number of
permutations of r things our of n things is denoted by npr.
Formulae:
1. n! = n(n-1)(n-2)....(1)
2. npr = n!/(n-r)!
Example: 10P4 = 10 x 9 x 8 x 7 = 5040
3. The number of ways in which n objects can be arranged in a circle is (n-
1)!
Combinations:
The ways in which a specified number of objects can be taken out of a
given number of objects (without regard to their arrangements) are called
Combinations. The symbol nCr denotes the number of combinations or r
things out of n things. Thus, for example the number of combinations of 2
objects out of three given objects a, b and c is 3, viz. ab, ca, bc.
Formulae:
1. nCr = nPr/r!
2. nCr = nCn-r
3. nCr + nCr-1 = (n+1)Cr
4. nC0 + nC1 + nC2 + nC3 + ......nCn = 2n
Circular Permutations
There are two types of circular permutations.
(a) If clockwise and anti clock-wise orders are different, then total number
of circular-permutations is given by (n-1)!
(b) If clock-wise and anti-clock-wise orders are taken as not different, then
total number of circular-permutations is given by (n-1)!/2!
Number of circular-permutations of n different things taken r at a time:-
(a) If clock-wise and anti-clockwise orders are taken as different, then
total number of circular-permutations = nPr /r
(b) If clock-wise and anti-clockwise orders are taken as not different, then
total number of circular permutation = nPr/2r
Examples
Example 1: In how many ways can the letters of the word ABACUS be
rearranged such that the vowels always appear together?
Solution: ABACUS is a 6 letter word with 3 of the letters being vowels.
If the 3 vowels have to appear together, then there will 3 other
consonants and a set of 3 vowels together.
These 4 elements can be rearranged in 4! Ways.
The 3 vowels can rearrange amongst themselves in 3!/2! ways as "a"
appears twice.
Hence, the total number of rearrangements in which the vowels appear
together are (4! x 3!)/2!
Example 2: How many different four letter words can be formed (the
words need not be meaningful) using the letters of the word
"MEDITERRANEAN" such that the first letter is E and the last letter is R?
Solution: The first letter is E and the last one is R.
Therefore, one has to find two more letters from the remaining 11 letters.
Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the
remaining 5 letters.
The second and third positions can either have two different letters or
have both the letters to be the same.
Case 1: When the two letters are different. One has to choose two
different letters from the 8 available different choices. This can be done in
8 * 7 = 56 ways.
Case 2: When the two letters are same. There are 3 options - the three
can be either Ns or Es or As. Therefore, 3 ways.
Total number of posssibilities = 56 + 3 = 59.
Example 3: How many different signals can be made by 5 flags from 8-
flags of different colours?
Solution: Number of ways taking 5 flags out of 8-flags = 8P5
= 8!/(8-5)!
= 8 x 7 x 6 x 5 x 4 = 6720
Example 4: A child has 3 pocket and 4 coins. In how many ways can he
put the coins in his pocket?
Solution: First coin can be put in 3 ways, similarly second, third and forth
coins also can be put in 3 ways.
So total number of ways = 3 x 3 x 3 x 3 = 34 = 81
Assignment
Discuss and post solution
1. A man has nine friends, four boys and five girls. In how many ways can
he invite them, if there have to be exactly three girls in the invitees?
2. A company manufactures pencils in boxes of 6, 9, and 20. The boxes
are sealed and the pencils cannot be sold loose. What is the largest
number of pencils that a wholesaler cannot purchase using some
combination of these boxes?
3. The number of ways in which 10 candidates A1, A2, ……, A10 can be
ranked so that A1 is always above A2 is ?
4. A hosted a party and invited all her friends and asked them to invite
their friends.There are n people in the party.Only S is not known to A.Each
pair that does not include A or S has exactly 2 common friends.Also,S
knows everyone except A.If only 2 friends can dance at a time,how many
dance numbers will be there at the party?
5. In a classroom there are 14 students seated in 3 rows of 5 chairs. The
place at the centre of the room is unoccupied. A teacher decides to
reassign the seats such that each student will occupy a chair adjacent to
his/her present one (i.e. move one desk forward, backward, right or left).
In how many ways can this reassignment be done?
6. Consider a 4 digit number. the first 2 digits are equal and last 2 digits
are also equal. How many of such digits are perfect square?
7. Suppose u have a currency, named x, in 3 denominations, 1, 10 and 50.
In how many ways can 107 x be given in this currency?
ProbabilityTable of ContentsTutorial Examples Assignment
Tutorial
An experiment is an act for which the outcome is uncertain. Examples of
experiments are rolling a die, tossing a coin, surveying a group of people
on their favorite soft drink, etc...
An experiment is said to be a random experiment, if it's out-come can't
be predicted with certainty.
Example; If a coin is tossed, we can't say, whether head or tail will appear.
So it is a random experiment.
A sample space S for an experiment is the set of all possible outcomes
of the experiment such that each outcome corresponds to exactly one
element in S. The elements of S are called sample points. If there is a
finite number of sample points, that number is denoted n(S), and S is said
to be a finite sample space.
For example, if our experiment is rolling a single die, the sample space
would be S = {1, 2, 3, 4, 5, 6}. If our experiment is tossing a single coin,
our sample space would be S = {Heads, Tails}.
Every subset of a sample space is an event. It is denoted by 'E'. e.g. In
throwing a dice S={1,2,3,4,5,6}, the appearance of an event number will
be the event E={2,4,6}.
An event, consisting of a single sample point is called a simple event.
e.g. In throwing a dice, S={1,2,3,4,5,6}, so each of {1},{2},{3},{4},{5}
and {6} are simple events.
Compound event: A subset of the sample space, which has more than
on element is called a mixed event.e.g. In throwing a dice, the event of
appearing of odd numbers is a compound event, because E={1,3,5}
which has '3' elements.
Equally likely events: Events are said to be equally likely, if we have no
reason to believe that one is more likely to occur than the other. e.g.
When a dice is thrown, all the six faces {1,2,3,4,5,6} are equally likely to
come up.
Exhaustive events: When every possible out come of an experiment is
considered. e.g. A dice is thrown, cases 1,2,3,4,5,6 form an exhaustive set
of events.
Probability of an Event
If 'S' be the sample space, then the probability of occurrence of an event
'E' is defined as:
P(E) = n(E)/N(S) = (number of elements in 'E'/ (number of elements in
sample space 'S')
Empirical Probability
Finding the probability of an empirical event is specifically based on direct
observations or experiences.
For example, a survey may have been taken by a group of people. If the
data collected is used to find the probability of an event tied to the
survey, it would be an empirical probability. Or if a scientist did research
on a topic and recorded the outcome and the data from this is used to find
the probability of an event tied to the research, it would also be an
empirical probability.
Equiprobable space
A sample space S is called an equiprobable space if and only if all the
simple events are equally likely to occur. e.g. A toss of a fair coin. It is
equally likely for a head to show up as it is for a tail.
Mutually Exclusive
Events E and F are said to be mutually exclusive if and only if they have
no elements in common.
E.g. if the sample space is rolling a die, where S = {1, 2, 3, 4, 5, 6}, and E
is the event of rolling an even number, E = {2, 4, 6} and F is the event of
rolling an odd number, F = {1, 3, 5}, E and F are mutually exclusive,
because they have NO elements in common.
Bayes' Theorem The short form of Bayes' Theorem states that if E and F are events,
then
P(F|E) =
P(E|F)P(F)
----------------------
P(E|F)P(F) + P(E|F')P(F')
Properties of Probability
1. 0 <= P(E) <= 1
2. P(not E) = 1 - P(E) So if, P(E) = 1/4 then P(not E) = 3/4.
3. "Or" probabilities with mutually exclusive events P (A or B) = P(A) +
P(B)
4. "Or" probabilities with events that are NOT mutually exclusive P (A or
B) = P(A) + P(B) - P(A AND B)
5. A and B are Independent Events if an only if P(A AND B) = P(A)P(B)
Examples
Example 1: Find the probability of getting a tail in tossing of a coin.
Solution: Sample space S = {H,T} and n(s) = 2
Event 'E' = {T} and n(E) = 1
therefore P(E) = n(E)/n(S) = 1/2
Example 2: A glass jar contains 6 red, 5 green, 8 blue and 3 yellow
marbles. If a single marble is chosen at random from the jar, what is the
probability of choosing a red marble? a green marble? a blue marble? a
yellow marble?
Solution: Outcomes: The possible outcomes of this experiment are
red, green, blue and yellow.
Probabilities:
P(red) = number of ways to choose red/total number of marbles =
6/22 = 3/11
P(green) = number of ways to choose green/total number of
marbles = 5/22
P(blue) = number of ways to choose blue/total number of marbles =
8/22 = 4/11
P(yellow) = number of ways to choose yellow/total number of
marbles = 3/22
Example 3: A man can hit a target once in 4 shots. If he fires 4 shots in
succession, what is the probability that he will hit his target?
Solution: The man will hit the target even if he hits it once or twice or
thrice or all four times in the four shots that he takes.
So, the only case where the man will not hit the target is when he fails to
hit the target even in one of the four shots that he takes.
The probability that he will not hit the target in one shot = 1 - 1/4 = 3/4
Therefore, the probability that he will not hit the target in all the four
shots =3/4 x 3/4 x 3/4 x 3/4 = 81/256
Hence, the probability that he will hit the target at least in one of the four
shots = 1 - 81/256
= 175/256 .
Example 4: What is the probability that the position in which the
consonants appear remain unchanged when the letters of the word Math
are re-arranged?
Solution: The total number of ways in which the word Math can be re-
arranged = 4! = 4*3*2*1 = 24 ways.
Now, if the positions in which the consonants appear do not change, the
first, third and the fourth positions are reserved for consonants and the
vowel A remains at the second position.
The consonants M, T and H can be re-arranged in the first, third and fourth
positions in 3! = 6 ways without the positions in which the positions in
which the consonants appear changing.
Therefore, the required probability = 3!/4! = 6/24 = 1/4
Assignment
Discuss and post solutions
1. Out of two-thirds of the total number of basket-ball matches, a team
has won 17 matches and lost 3 of them. What is the maximum number of
matches that the team can lose and still win three-fourths of the total
number of matches, if it is true that no match can end in a tie?
2. 4 people played a game of chess, where each one plays every other
player. What is the maximum number of points that any player could
gather if every win gets him 1 point ?
3. From a pack of 52 cards, all face cards are removed and four cards are
drawn. Then the probability that they are of different suit and different
denomination is
4. Five balls of different colours are to be placed in three boxes of
different sizes. Each box can hold all five balls. The number of ways in
which we can place the balls in the boxes so that no box remains empty is
?
5. How many arrangements can be made of the letters of the word
DRAUGHT the vowels never being separated?
6. If the integers m and n are chosen at random from integers from
integers 1 to 100 with replacement, then the probability that a no. of the
form 7m + 7n is divisible by 5 equals?
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