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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


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


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


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


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


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


= = 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


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


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



LEARNING

OUTCOMES : At the end of the lesson students should be able to:

9

Taxi Bus

Penang Train KL Taxi Bus

Genting Highland


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


=

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


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


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


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


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


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)


or

The number of permutations = 2 ( 1 6 5 4 )

= 240 ways

c)


d)


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


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


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



LEARNING


a) Find the number of permutations of n objects

comprising of r1 identical objects, r2 identical objects,…, rk

identical objects.

19


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


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


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


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


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


SUBTOPIC : 9.2 Combinations

LEARNING


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


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


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


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


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


= 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


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


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


Answers

1. 7560

2. 21

3. (a) 15 (b) 30

32

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