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Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
LECTURE 1 0F 4
TOPIC : 9.0 PERMUTATIONS AND COMBINATIONS
SUBTOPIC : 9.1 Permutations
LEARNING
OUTCOMES : At the end of the lesson students should be able to :
a) Understand the techniques of counting.
b) Understand permutation of a set of objects.
c) Find the number of permutations of n different objects.
Permutations of a set of objects
Introduction
A permutation of a set of objects is any arrangement of those objects in a definite order
(order is important).
Suppose there are 4 ways from Johor to Penang and 2 ways from Penang to Pulau
Langkawi. How many ways can we go for a journey from Johor to Pulau Langkawi through
Penang ?
Bus
Taxi Flight
Johor Train Penang Ferry P.Langkawi
Van
1
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
Students will count or list all the 8 possible ways.
Answer : We can use permutation to solve this problem. The total number of possible ways is
.
Consider another example, if A= , then ab is a two-element permutation of A, acd is a
three-element permutation of A, and adcb is a four-element permutation of A. The order in
which objects are arranged is important. For example, ab and ba are considered different two-
element permutations, abc and cba are distinct three-element permutations, and abcd and cbad
are different four-element permutations.
For another example the six permutations of ABC are the six different arrangements of ABC.
These are
ABC ACB BAC BCA CAB CBA
The number of permutations can be calculated using the multiplication principle.
Example
A fair coin and a die are tossed together. How many different outcomes are possible ?
Solution
The coin has two possible outcomes (head, H and Tail, T) and the dice has 6 possible outcomes.
The number of different possible outcomes is ___ x ___ = ____
Multiplication Principle
If there are m ways for an event to occur and n ways for another event to
occur, then there are m n ways for the two events to occur.
2
2 6 12
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
Die
1 2 3 4 5 6C
oin Head (H) (H,1) (H ,2) (H, 3) (H, 4) (H, 5) (H, 6)
Tail (T) (T,1) (T,2) (T, 3) (T, 4) (T, 5) (T, 6)
Table 9.1.1
From the table 9.1.1, the possible outcomes are
____ ____ ____ ____ ____ ____
____ ____ ____ ____ ____ ____
Example
A shop stocks T-shirts in four sizes : small, medium and large. They are available in four
colours; black , red , yellow and green. If the sizes are denoted by S, M and L and the colours
are denoted by B, R, Y and G make a list of all the different labels needed to distinguish the T-
shirts and find the number of different labels.
Solution
3
(H,1) (H,2) (H,3)
(T,2) (T,3) (T,5)
(H,4)
(T,4)(T,1)
(H,6)(H,5)
(T,6)
S
M
L
GYRB
GYRB
GYRB
SBSRSYSGMBMRMYMGLBLRLYLG
Sizes Colours Outcomes
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
A list of all the different labels are
SB SR SY SG MB MR MY MG LB LR LY LG
The number of different labels is 3 x 4 = 12
Permutations of n different objects
Let us consider the number of ways of arranging n letters. If we have 1 letter, there is just one
arrangement, eg: A. If we have 2 letters, there are two different arrangements, eg : AB and BA.
If we have 3 letters, the different arrangements of the letters A, B and C are :
The first letter can be
there are three ways of choosing the first letter.
When the first letter has been chosen, there are two letters from which to choose the second; and
the possible ways of choosing the first two letters are:
there are two ways of choosing the second letter
i.e. for each of the three ways of choosing the first letter, there are two ways of choosing the
second letter. Hence there are 3 2 ways of choosing the first two letters. Having chosen the
first two letters, there is only one choice for the third letter, i.e. for each of the 3 2 ways of
choosing the first two letters, there is only one possibility for the third letter. Hence there are 3
2 1 ways of arranging the three letters A, B and C.
Now considering another example,
how many different ways do you think there are of arranging 4 letters?
4
SB
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
You should able to see there will be 24 different ways, which is found from 4 3 2 1. If
there are 500 different objects, the number of ways would be 500 499 498 … 3 2 1.
This is tedious to write, so we use the notation 500! ( 500 factorial )
In general,
Notes :
means the products of all the integers from 1 to n inclusive and is called
‘n factorial’.
Example
List the set of all permutations of the symbols P, Q and R when they are taken 3 at a time.
Solution
PQR, PRQ, QPR, QRP, RPQ, RQP i.e. 6 of them.
Example
How many three-digit numbers can be made from the integers 2, 3, 4 ?
Solution
Number of permutations of n different objects taken all at a time without repetition
n (n – 1) (n – 2) …… 2 1
= n !
5
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
n = 3
= = 3 ! = 3 2 1 = 6
The number of arrangements is 6.
Example
In how many ways can ten instructors be assigned to ten sections of a course in mathematics?
Solution
Substituting n = 10, we get
= = 10 ! = 3,628,800 ways
Example
Three people, Aishah, Badrul and Daniel must be scheduled for job interviews. In how many
different orders can this be done?
Solution
n = 3
So there are 3! = 6 possible orders for the interviews.
Example
How many different 4 digit numbers can be formed from the digits 5, 6, 7 and 8
i) if no repetitions
n = 4
6
4 possible choices
3 possible choices
2 possible choices
1 possible choice
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
= = 4! = 24
ii) if the first digit must be 7 and no repeatation.
7
= 1! 3! = 6
Note : If repetition is allowed, we can choose from all 2 digits for each digit of the
number (a digit can be used more than once).
How many different ways of arranging 3 digit numbers from digits 5 and 6 ?
7
Only 1 possible choice
3 possible choices
2 possible choices
1 possible choice
5
6
5
6
5
6
5
65
65
65
6
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
You should be able to see there will be 8 different ways, which is found from 2 2 2.
Exercise
1. How many permutations of the set begin with a and end with c ?
2. Six people are going on a motoring holiday in a six-seater car. In how many ways can
they be seated if all six are able to drive?
8
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
3. If there are 3 ways from Penang to Kuala Lumpur and 2 ways from Kuala Lumpur to
Genting Highlands, how many ways can we go for a journey from Penang to Genting
Highlands through Kuala Lumpur ?
Answers
1. 6 2. 720 3. 6
LECTURE 2 OF 4
TOPIC : 9.0 PERMUTATIONS AND COMBINATIONS
SUBTOPIC : 9.1 Permutations
LEARNING
OUTCOMES : At the end of the lesson students should be able to:
9
Taxi Bus
Penang Train KL Taxi Bus
Genting Highland
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
a) Find the number of permutations of r objects from n
different objects.
CONTENT
Permutations of r objects from n different objects
a) With restriction ( )
b) With no restriction ( )
c) Permutation with conditions
Permutations of r objects taken from n different objects
A permutation of r objects taken from n different objects without repetition is an arrangement of
the objects in a specific order.
For example, there 12 permutations for the letters A, B, C and D taken 2 at a time.
These are : AB BA CA DA
AC BC CB DB
AD BD CD DC
Using the multiplication principle, the number of permutations of 4 objects taken two at a time =
4 3 = 12. Similarly, the number of permutations of 10 objects taken 3 at a time =10 9 8 =
720.
In general,
the number of permutations of n objects taken r at a time
=
=
10
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
=
The number of permutations of r objects chosen from a set of n different objects is
denoted by
Example
Suppose you have 4 different flags. How many different signals could you make using
(i) 2 flags
(ii) 2 or 3 flags
Solution
(i) n = 4 r = 2
There are 12 different signals using 2 flags from 4 flags.
(ii) n = 4, r = 2 or n = 4, r = 3
There are 36 different signals using 2 or 3 flags from 4 flags.
Example
How many arrangements of the letters of the word B E G I N are there if
(i) 3 letters are used
(ii) all of the letters are used
11
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
Solution
(i) n = 5 r = 3
The arrangements of the letters of the word B E G I N is 60 if 3 letters are used
(ii) n = 5
Number of arrangements if all of the letters are used.
=
Example
A relay team has 5 members. How many ways can a coach arrange 4 of them to run a 4x100 m
race
Solution
The order of the four runners is important.
Number of arrangements the coach can make =
= 120
Permutations with condition
Example How many three-digit numbers can be made from the integers 2, 3, 4, 5, 6 if
(i) each integer is used only once?
(ii) there is no restriction on the number of times each integer can be used?
12
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
Solution
(i) n = 5 r = 3
There are 60 different arrangements.
(ii) Number of ways of making the three-digit numbers
= 5 5 5 (repetition is allowed)
= 125
Example
Find the number of arrangements of 4 digits taken from the set { 1, 2, 3, 4}
In how many ways can these numbers be arranged so that
(a) The numbers begin with digit ‘1’
(b) The numbers do not begin with digit ‘1’
Solution
Number of arrangements of 4 digits = 4! = 24
(a) If the arrangements begin with digit ‘1’, then the number of ways the 3 remaining
digits can be arranged = 3! = 6
(b) The number of arrangements that do not begin with digit ‘1’ = 24 – 6
= 18
Example
Four sisters and two brothers are arranged in different ways in a straight line for several
photographs to be taken. How many different arrangements are possible if
(a) there are no restrictions
(b) the two brothers must be separated
13
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
Solution
(a) Number of arrangements of 6 people = 6!
= 720
(b) First, find the numbers of arrangements with the two brothers standing next to each
other. In these arrangements, the two brothers move together as one unit and this is
equivalent to the arrangement of 5 objects except that they are able to switch positions with
each other.
Number of arrangements with two brothers next to each other = 5! 2!
= 120 2
= 240
Number of arrangements with the two brothers separated = 720 – 240
= 480
Example
Arrange 6 boys and 3 girls in a straight line so that the girls are separated. In how many ways can
this be done ?
Solution
Consider this arrangement : B B B B B B
14
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
Let the 6 B’s represent the 6 boys and the ‘ ’ represent the spaces for the girls.
Number of arrangements for the boys = 6!
Number of arrangements for the girls = (7 spaces available for the 3 girls)
= 210
Total number of arrangements of 6 boys and 3 girls where the girls are separated
= 6! 210
= 151200
Example
There are 10 students out of whom six are females. How many possible arrangements are
there if
a) they are arranged in a row?
b) males always sit on one side and female on the other side?
Solution
a) The number of permutations = 10! = 3628800
b) 2!
6! 4!
6 female 4 male
The number of permutations = 2! 6! 4! = 34560
Example
A witness to a hit-and-run accident told the police that the plat number contained the
letters PDW followed by 3 digits, the first of which is 5. If the witness cannot recall the
last 2 digits, but is certain that all 3 digits are different, find the maximum number of
automobile registrations that the police may have to check.
15
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
Solution
P D W
The number of permutations = 1 9 8 = 72 ways
Example
In how many ways can 4 girls and 5 boys sit in a row if the boys and girls must sit
alternate to each other?
Solution
B B B B B
The number of permutations = 5! 4! = 2880 ways
Example
Four digit numbers are to be formed from the digits 0, 1, 2, 3, 4, 5, 6 without repetition .
How many numbers can be formed if each number
a) is less than 5000
b) begins with digit 4 or 6
c) is between 2000 and 6000
d) is an odd number
Solution
a)
The number of permutations = 4 6 5 4 = 480 ways
b)
16
4 6
5
4 choices (1, 2, 3, 4)
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
or
The number of permutations = 2 ( 1 6 5 4 )
= 240 ways
c)
The number of permutations = 4 6 5 4 = 480 ways
d)
The number of permutations = 5 5 4 3 = 300 ways
Example
How many four-digit even numbers can be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7 to
make up numbers between 2000 and 6000
a) without repetition
b) with repetition
Solution
a) consider the last position by two parts : “0” and “not 0”
ends with 0 or not ends with 0
The number of permutations = (4 6 5 1 ) + ( 3 6 5 3) = 390 ways
b)
17
0
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
The number of permutations = 4 8 8 4 = 1024 ways.
Example
Three married couples have bought 6 seats in the same row for a concert. In how many
different ways can they be seated
a) with no restrictions
b) if each couple is to sit together
c) if all the men sit together to the right of all the women
Solution
a)
The number of permutations = 6! = 720
b) H1 W1 H2 W2 H3 W3
The number of permutations = 3! 2! 2! 2! = 48 ways
c) W1 W2 W3 H1H2H3
The number of permutations = 1 3! 3! = 36 ways.
Exercise
1. Find the number of different ways in which a gold, a silver and a bronze medal can be
awarded to 15 competitors if each competitor can win only one medal.
2. In how many different ways may 10 different letters be placed in 15 different boxes, not
more than one letter being placed in any box? You may leave the answer in factorial form.
3. A shop has 5 different printers but there is space for only 3 printers on the display
H1 W1 H2 W2 H3 W3
18
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
shelf. How many arrangements are possible ?
4. A photographer wishes to arrange 7 children consisting of 3 boys and 4 girls in a
straight line for a picture. In how many ways can he do this if
(a) the girls are separated
(b) the 3 boys occupy the 3 central positions
5. Find the number of ways ABCDE can be arranged if
(a) the arrangements must begin with the letter A.
(b) do not begin with the letter A
Answers
1. 2730
2.
3. 60
4. (a) 144 (b) 144
5. (a) 24 (b) 96
LECTURE 3OF 4
TOPIC : 9.0 PERMUTATIONS AND COMBINATIONS
SUBTOPIC : 9.1 Permutations
LEARNING
OUTCOMES : At the end of the lesson students should be able to:
a) Find the number of permutations of n objects
comprising of r1 identical objects, r2 identical objects,…, rk
identical objects.
19
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
Permutations where some objects are repeated
Consider the number of permutations of the letters D E F E A T E D.
There are three E’s and two D’s. The number of permutations of D1 E1 F E2 A T E3 D2 is
8! But E1, E2, E3 can be arranged in 3! ways, and D1 D2 can be arranged in 2! ways, so
The number of arrangements of D1 E1 F E2 A T E3 D2 is 3! x 2! times the number of
arrangements of D E F E A T E D.
Therefore the number of permutations of D E F E A T E D is .
Generalising this argument we see that,
The number of permutations of n objects comprising of r1 identical objects, r2 identical
objects, ………., rk identical objects is
Example
How many different permutations can be made using the letters of the words
(i) BOOKS (ii) LOTTO (iii) MATHEMATICS
Solution
(i) n = 5 r1 = 2
The number of different permutations is 60.
(ii) n = 5 r1 = 2 r2 = 2
20
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
There are 30 different arrangement.
(iii) n = 11 r1 = 2 r2 = 2 r3 = 2
= 4989600
Example
There are 2 copies of each of 3 different books to be arranged on a shelf. In how many
distinguishable ways can this be done?
Solution
n = 2 3 = 6 ( there are six books )
r1 = 2 r2 = 2 r3 = 2
There are 90 ways to arrange 2 copies of each of 3 different books on a shelf.
Example
How many different 10-letter codes can be made using three a’s, four b’s, and three c’s?
Solution
n = 10 r1 = 3 r2 = 4 r3 = 3
The number of such codes is
= 4200.
21
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
Example:
In how many ways can 3 red, 4 blue and 2 green pens be distributed among nine students seated
in a row if each student receives one pen?
Solution
The number of permutations = = 1260 ways
Example
In how many of the possible permutations of the letters of the word ADDING are the two D’s:
(i) together, (ii) separated
Solution
(i) There are five different items ( A, (DD), I, N, G ) which can be arranged in 5! ways.
The number of possible permutations is 5! = 5 4 3 2 1 = 120.
(ii) (number of arrangements with D’s separated) = (number of arrangements without
restriction) - (number of arrangements with D’s together).
Now the number of arrangements without restriction is .
The number of arrangements in which the D’s are separated is
- 5! = ( 6 )( 5 )( 4 )( 3 ) – 120 = 240.
Example
How many different arrangements are there for the letters of the word
ARRANGEMENTS if
22
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
a) begins with “R” and ends with “E”
b) the two letters “E” are separated
c) the two letters “E” and the two letters “A” are together
d) the consonant letters GMTS are together
e) the two letters “N” occupied both ends
Solution
a)
The number of permutations = (2! for A, 2! for N)
= 907200
b) The number of permutations separated
= without restriction – E together
= _ EE
= _ (2! for A, 2! for R, 2! for N)
= 24948000
c) EE AA
The number of permutations = 1 1 = 907200
d) GMTS
The number of permutations = 4! = 544320
e)
23
R E
10 letters
10 letters
8 letters
8 letters
N N
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
The number of permutations = 1 = 453600
Exercise
1. A dancing contest has 11 competitors, of whom three are Americans, two are Malaysians,
three are Indonesians, and three are Italians. If the contest result lists only the nationality of
the dancers, how many outcomes are possible?
2. In how many ways can the letters of the word STATISTICS be arranged?
Answers
1. 92,400 2. 50400
LECTURE 4OF 4
TOPIC : 9.0 PERMUTATIONS AND COMBINATIONS
SUBTOPIC : 9.2 Combinations
LEARNING
OUTCOMES : At the end of the lesson students should be able to:
a) Understand combination of a set of object
b) Determine the number of ways to form combinations of r
objects from n objects.
Combinations of a set of objects
A combination is a selection of objects with no consideration given to the order (arrangement)
of the object.
24
10 letters
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
So while ABC and BCA are different permutations they are the same combination of letters.
PQR, PRQ, QPR, QRP, RPQ, RQP are considered as 1 combination (because the order is not
considered) and 6 permutations (because the order is considered).
Example:
Determine whether each of the following is a permutation or combination:
a) 5 pictures placed in a row. (Permutation)
b) 3 story books picked from a rack. (Combination)
c) A team of 9 players chosen from a group of 20. (Combination)
d) The arrangements of the letters in the word OCTOBER. (Permutation)
e) Types of food in a plate taken for lunch consist of rice, vegetables, chicken curry and
prawn paste sambal. (Combination)
Therefore, we can conclude that permutations are used when order is important and
combinations are used when order is not important.
Combinations of r object from n objects
Three students could be chosen, in order, from a total of 18 in ways. However, each of
these choices can be arranged in 3! ways (ABC, ACB, BAC, BCA, CAB, CBA), so the number
of combinations of three items chosen from 18 is = 816.
The general result is:
The number of combinations (or selections) of r objects chosen from n unlike objects is
Notes:
: is the number of combinations on n symbols taken r at a time.
25
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
Consider the following table;
N different
object
taken r
object
Combinatio
n Number of
Combinations
PermutationNumber of
permutations
A, B 2 AB 1 AB, BA 2
A, B, C 2 AB, AC, BC 3
AB, BA, AC, CA,
BC, CB 6
A, B, C 3 ABC 1ABC, ACB, BCA,
BAC, CAB, CBA6
From the table,
The number of combinations of 2 objects A, B taken 2 at a time = 2C2 = 1.
The number of combinations of 3 objects A, B, C taken 2 at a time = 3C2 = = 3.
The number of combinations of 3 objects A, B, C taken 3 at a time = 3C3 = 1.
Thus, we can see that the number of combinations is always less than the number of
permutations.
Example
A quiz team of four is chosen from a group of 15 students. In how many ways could the team be
chosen?
Solution
26
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
Therefore the team can be chosen in 1365 ways.
Example
If there are eight girls and seven boys in a class, in how many ways could a group be chosen so
that there are two boys and two girls in the group?
Solution
Two girls can be chosen in ways, and the two boys can be chosen in ways. For each of
the ways of selecting the girls there are ways of selecting the boys.
Using the multiplication principle , number of ways of selecting the group.
=
= 588
So there are 588 ways of choosing a group consisting of two boys and two girls.
Example
A school committee consists of six girls and four boys. A social sub-committee consisting of
four students is to be formed. In how many ways could the group be chosen if there are to be
more girls than boys in the group?
Solution
If there are to be more girls than boys in the group then the group will either have four girls and
no boys, or three girls and one boy.
Four girls can be chosen in ways = 15 ways
Three girls and one boy can be chosen in = = 20 4 = 80 ways
Therefore the number of ways of choosing the group if there are more girls than boys is
15 + 80 = 95 ways
27
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
Example
Given the set S = {a, b, c, d, e} consists of 5 elements. List all the subsets of S with (a) two elements (b) four elements
Solution (a) {a, b}, {a, c}, {a, d}, {a, e}, {b, c}, {b, d}, {b, e}, {c, d}, {c, e}, {d, e}
The list show combinations of 5 elements taken 2 at a time.
(b) {a, b, c, d}, {a, b, c, e}, {a, b, d, e}, {a, c, d, e}, {b, c, d, e}
The list shows combinations of 5 elements taken 4 at a time.
Note : {a, b, c, d} and {c, a, b, d} are the same subset because they contain
exactly the elements and are considered as one (similar) combination.
Example
In a football training squad of 24 people, 3 are goalkeepers, 7 are defenders, 6 are midfielders
and 8 are forwards. A final squad of 16 selected for a match must consist of 2 goalkeepers, 4
defenders, 5 midfielders and 5 forwards. Find the number of possible selections if one particular
goalkeeper, 2 particular defenders, 3 particular midfielders and 3 particular forwards are
automatically selected.
Solution
Number of ways of selecting the goalkeepers = 1C1 = 2
Number of ways of selecting the defenders = 2C2 = 10
Number of ways of selecting the midfielders = 3C3 = 3
Number of ways of selecting the forwards = 3C3 = 10
Number of ways of selecting the squad
= 2 10 3 10 (by using the principle of multiplication)
28
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
= 600
Example
ABCDEFGH is a regular octagon.
(a) How many triangles can be formed with the vertices of the octagon as vertices ?
(b) How many diagonals can be drawn by joining the vertices?
Solution
(a) A triangle is formed by taking 3 of the vertices
Number of triangles =
= 56
(b) A line can be formed by taking any 2 points from the 8 vertices of the octagon
Number of lines formed = = 28
These 28 lines include the 8 sides of the of the octagon
Thus the number of diagonals = 28 – 8
= 20
Example15 students are divided into 3 groups, with A having 7 students, group B having 5 students and
group C having 3 students. Find the number of ways to form
a) the 3 groups
29
A
G
F
E
D
C
BH
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
b) the 3 groups with 2 given students must be in group A.
Solutiona) The number of combinations = 15C7 8C5 3C3 = 360360
b) The number of combinations = (2C2 13C5 ) 8C5 3C3 = 72072
ExampleA 3 member committee is to be formed from 4 couples of husband and wife. Find the possible
number of committees that can be formed if
a) all the members are men
b) the husband and the wife cannot be in the committee at the same time.
Solutiona) The number of combinations = 4C3 = 4 ways
b) The number of selecting 3 groups from 4 groups = 4C3
The number of choosing a person from each of the 3 groups selected
= 2C1 2C1 2C1
= 8
Thus, the number of combinations = 4C3 2C1 2C1 2C1 = 32 ways
ExampleIn a test, a candidate is required to answer 8 out of 10 questions. Find the number of ways a
candidates
a) can answer the questions
b) can answer the question if the first 3 questions must be answered.
30
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
Solution
a) 10C8 = 45 b) 7C5 = 21
ExampleIn how many ways can a teacher choose one or more students as a prefects from 5 eligible
students?
Solution
The number of combinations is may be one or two or three or four or five person chosen
= 5C1 + 5C2 + 5C3 + 5C4 + 5C5
= 66
Exercise
1. A father buys nine different toys for his four children. In how many ways can he give one
child three toys and the remaining three children two toys each?
2. A party of nine people consists of five men and four women, and a group of four people is to
be chosen at random from this party. In a how many ways can a group of four be chosen that
contains at least three women?
3. For a badminton doubles game, 2 players are chosen from among 5 male players and
3 female players to represent a club. In how many ways can this doubles pair be
selected if
(a) the team is a mixed double, comprising one male player and one female player?
(b) The team is either a male pair or a female pair and no mixed pair are allowed?
31
Mathematics QS026Topic 9 : Permutations and Combinations – Lesson Plan
Answers
1. 7560
2. 21
3. (a) 15 (b) 30
32