Jacob Bernoulli

(1654-1705)

vEvery body of discovery is mathematical in form because there is no

other guidance we can have – DARWINv

7.1 Introduction

Suppose you have a suitcase with a number lock. The number

lock has 4 wheels each labelled with 10 digits from 0 to 9.

The lock can be opened if 4 specific digits are arranged in a

particular sequence with no repetition. Some how, you have

forgotten this specific sequence of digits. You remember only

the first digit which is 7. In order to open the lock, how

many sequences of 3-digits you may have to check with? To

answer this question, you may, immediately, start listing all

possible arrangements of 9 remaining digits taken 3 at a

time. But, this method will be tedious, because the number

of possible sequences may be large. Here, in this Chapter,

we shall learn some basic counting techniques which will enable us to answer this

question without actually listing 3-digit arrangements. In fact, these techniques will be

useful in determining the number of different ways of arranging and selecting objects

without actually listing them. As a first step, we shall examine a principle which is most

fundamental to the learning of these techniques.

7.2 Fundamental Principle of Counting

Let us consider the following problem. Mohan has 3 pants and 2 shirts. How many

different pairs of a pant and a shirt, can he dress up with? There are 3 ways in which

a pant can be chosen, because there are 3 pants available. Similarly, a shirt can be

chosen in 2 ways. For every choice of a pant, there are 2 choices of a shirt. Therefore,

there are 3 × 2 = 6 pairs of a pant and a shirt.

7

Chapter

PERMUTATIONS AND COMBINATIONS

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PERMUTATIONS AND COMBINATIONS 135

Let us name the three pants as P

1

, P

2

, P

3

and the two shirts as S

1

, S

2

. Then,

these six possibilities can be illustrated in the Fig. 7.1.

Let us consider another problem

of the same type.

Sabnam has 2 school bags, 3 tiffin boxes

and 2 water bottles. In how many ways

can she carry these items (choosing one

each).

A school bag can be chosen in 2

different ways. After a school bag is

chosen, a tiffin box can be chosen in 3

different ways. Hence, there are

2 × 3 = 6 pairs of school bag and a tiffin

box. For each of these pairs a water

bottle can be chosen in 2 different ways.

Hence, there are 6 × 2 = 12 different ways in which, Sabnam can carry these items to

school. If we name the 2 school bags as B

1

, B

2

, the three tiffin boxes as T

1

, T

2

, T

3

and

the two water bottles as W

1

, W

2

, these possibilities can be illustrated in the Fig. 7.2.

Fig 7.1

Fig 7.2

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136 MATHEMATICS

In fact, the problems of the above types are solved by applying the following

principle known as the fundamental principle of counting, or, simply, the multiplication

principle, which states that

“If an event can occur in m different ways, following which another event

can occur in n different ways, then the total number of occurrence of the events

in the given order is m×n.”

The above principle can be generalised for any finite number of events. For

example, for 3 events, the principle is as follows:

‘If an event can occur in m different ways, following which another event can

occur in n different ways, following which a third event can occur in p different ways,

then the total number of occurrence to ‘the events in the given order is m × n × p.”

In the first problem, the required number of ways of wearing a pant and a shirt

was the number of different ways of the occurence of the following events in succession:

(i) the event of choosing a pant

(ii) the event of choosing a shirt.

In the second problem, the required number of ways was the number of different

ways of the occurence of the following events in succession:

(i) the event of choosing a school bag

(ii) the event of choosing a tiffin box

(iii) the event of choosing a water bottle.

Here, in both the cases, the events in each problem could occur in various possible

orders. But, we have to choose any one of the possible orders and count the number of

different ways of the occurence of the events in this chosen order.

Example 1

Find the number of 4 letter words, with or without meaning, which can be

formed out of the letters of the word ROSE, where the repetition of the letters is not

allowed.

Solution

There are as many words as there are ways of filling in 4 vacant places

by the 4 letters, keeping in mind that the repetition is not allowed. The

first place can be filled in 4 different ways by anyone of the 4 letters R,O,S,E. Following

which, the second place can be filled in by anyone of the remaining 3 letters in 3

different ways, following which the third place can be filled in 2 different ways; following

which, the fourth place can be filled in 1 way. Thus, the number of ways in which the

4 places can be filled, by the multiplication principle, is 4 × 3 × 2 × 1 = 24. Hence, the

required number of words is 24.

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PERMUTATIONS AND COMBINATIONS 137

A

Note If the repetition of the letters was allowed, how many words can be formed?

One can easily understand that each of the 4 vacant places can be filled in succession

in 4 different ways. Hence, the required number of words = 4 × 4 × 4 × 4 = 256.

Example 2

Given 4 flags of different colours, how many different signals can be

generated, if a signal requires the use of 2 flags one below the other?

Solution

There will be as many signals as there are ways of filling in 2 vacant places

in succession by the 4 flags of different colours. The upper vacant place can

be filled in 4 different ways by anyone of the 4 flags; following which, the lower vacant

place can be filled in 3 different ways by anyone of the remaining 3 different flags.

Hence, by the multiplication principle, the required number of signals = 4 × 3 = 12.

Example 3 How many 2 digit even numbers can be formed from the digits

1, 2, 3, 4, 5 if the digits can be repeated?

Solution There will be as many ways as there are ways of filling 2 vacant places

in succession by the five given digits. Here, in this case, we start filling in unit’s

place, because the options for this place are 2 and 4 only and this can be done in 2

ways; following which the ten’s place can be filled by any of the 5 digits in 5 different

ways as the digits can be repeated. Therefore, by the multiplication principle, the required

number of two digits even numbers is 2 × 5, i.e., 10.

Example 4 Find the number of different signals that can be generated by arranging at

least 2 flags in order (one below the other) on a vertical staff, if five different flags are

available.

Solution A signal can consist of either 2 flags, 3 flags, 4 flags or 5 flags. Now, let us

count the possible number of signals consisting of 2 flags, 3 flags, 4 flags and 5 flags

separately and then add the respective numbers.

There will be as many 2 flag signals as there are ways of filling in 2 vacant places

in succession by the 5 flags available. By Multiplication rule, the number of

ways is 5 × 4 = 20.

Similarly, there will be as many 3 flag signals as there are ways of filling in 3

vacant places in succession by the 5 flags.

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138 MATHEMATICS

The number of ways is 5 × 4 × 3 = 60.

Continuing the same way, we find that

The number of 4 flag signals = 5 × 4 × 3 × 2 = 120

and the number of 5 flag signals = 5 × 4 × 3 × 2 × 1 = 120

Therefore, the required no of signals = 20 + 60 + 120 + 120 = 320.

EXERCISE 7.1

1. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5

assuming that

(i) repetition of the digits is allowed?

(ii) repetition of the digits is not allowed?

2. How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the

digits can be repeated?

3. How many 4-letter code can be formed using the first 10 letters of the English

alphabet, if no letter can be repeated?

4. How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if

each number starts with 67 and no digit appears more than once?

5. A coin is tossed 3 times and the outcomes are recorded. How many possible

outcomes are there?

6. Given 5 flags of different colours, how many different signals can be generated if

each signal requires the use of 2 flags, one below the other?

7.3 Permutations

In

Example 1 of the previous Section, we are actually counting the different possible

arrangements of the letters such as ROSE, REOS, ..., etc. Here, in this list, each

arrangement is different from other. In other words, the order of writing the letters is

important. Each arrangement is called a permutation of 4 different letters taken all

at a time. Now, if we have to determine the number of 3-letter words, with or without

meaning, which can be formed out of the letters of the word NUMBER, where the

repetition of the letters is not allowed, we need to count the arrangements NUM,

NMU, MUN, NUB, ..., etc. Here, we are counting the permutations of 6 different

letters taken 3 at a time. The required number of words = 6 × 5 × 4 = 120 (by using

multiplication principle).

If the repetition of the letters was allowed, the required number of words would

be 6 × 6 × 6 = 216.

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Definition 1

A permutation is an arrangement in a definite order of a number of

objects taken some or all at a time.

In the following sub-section, we shall obtain the formula needed to answer these

questions immediately.

7.3.1 Permutations when all the objects are distinct

Theorem 1

The number of permutations of n different objects taken r at a time,

where 0 < r ≤ n and the objects do not repeat is n ( n – 1) ( n – 2). . .(

n – r + 1),

which is denoted by

n

P

r

.

Proof There will be as many permutations as there are ways of filling in r vacant

places

. . . by

←

r vacant places

→

the n objects. The first place can be filled in n ways; following which, the second place

can be filled in (n – 1) ways, following which the third place can be filled in (n – 2)

ways,..., the rth place can be filled in (n – (r – 1)) ways. Therefore, the number of

ways of filling in r vacant places in succession is n(n – 1) (n – 2) . . . (n – (r – 1)) or

n ( n – 1) (n – 2) ... (n – r + 1)

This expression for

n

P

r

is cumbersome and we need a notation which will help to

reduce the size of this expression. The symbol n! (read as factorial n or n factorial )

comes to our rescue. In the following text we will learn what actually n! means.

7.3.2 Factorial notation The notation n! represents the product of first n natural

numbers, i.e., the product 1 × 2 × 3 × . . . × (n – 1) × n is denoted as n!. We read this

symbol as ‘n factorial’. Thus, 1 × 2 × 3 × 4 . . . × (n – 1) × n = n !

1 = 1 !

1 × 2 = 2 !

1× 2 × 3 = 3 !

1 × 2 × 3 × 4 = 4 ! and so on.

We define 0 ! = 1

We can write 5 ! = 5 × 4 ! = 5 × 4 × 3 ! = 5 × 4 × 3 × 2 !

= 5 × 4 × 3 × 2 × 1!

Clearly, for a natural number n

n ! = n (n –

1) !

= n (n – 1) (n – 2) ! [provided (n ≥ 2)]

= n (n – 1) (n – 2) (n – 3) ! [provided (n ≥ 3)]

and so on.

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140 MATHEMATICS

Example 5 Evaluate (i) 5 ! (ii) 7 ! (iii) 7 ! – 5!

Solution (i) 5 ! = 1 × 2 × 3 × 4 × 5 = 120

(ii) 7 ! = 1 × 2 × 3 × 4 × 5 × 6 ×7 = 5040

and (iii) 7 ! – 5! = 5040 – 120 = 4920.

Example 6 Compute (i)

7!

5!

(ii)

( )

12!

10! (2!)

Solution (i) We have

7!

5!

=

7 6 5!

5!

× ×

= 7 × 6 = 42

and (ii)

( ) ( )

12!

10! 2!

=

(

)

( ) ( )

12 11 10!

10! 2

× ×

×

= 6 × 11 = 66.

Example 7 Evaluate

( )

!

! !

n

r n r

−

, when n = 5, r = 2.

Solution We have to evaluate

( )

5!

2! 5 2 !

−

(since n = 5, r = 2)

We have

( )

5!

2 ! 5 2 !

−

=

5! 5 4

10

2! 3! 2

×

= =

×

.

Example 8 If

1 1

8! 9! 10!

x

+ =

, find x.

Solution We have

1 1

8! 9 8! 10 9 8!

x

+ =

× × ×

Therefore

1

1

9 10 9

x

+ =

×

or

10

9 10 9

x

=

×

So x = 100.

EXERCISE 7.2

1. Evaluate

(i) 8 ! (ii) 4 ! – 3 !

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PERMUTATIONS AND COMBINATIONS 141

2. Is 3 ! + 4 ! = 7 ! ? 3. Compute

8!

6! 2!

×

4. If

1 1

6! 7! 8!

x

+ =

, find x

5. Evaluate

( )

!

!

n

n r

−

, when

(i) n = 6, r = 2 (ii) n = 9, r = 5.

7.3.3 Derivation of the formula for

n

P

r

( )

!

P

!

n

r

n

n r

−

=

, 0 ≤ r ≤ n

Let us now go back to the stage where we had determined the following formula:

n

P

r

= n (n – 1) (n – 2) . . . (n – r + 1)

Multiplying numerator and denomirator by (n – r) (n – r – 1) . . . 3 × 2 × 1, we get

(

)

(

)

(

)

(

)

(

)

( )( )

1 2 1 1 3 2 1

P

1 3 2 1

n

r

n n n ... n r n r n r ...

n r n r ...

− − − + − − − × ×

=

− − − × ×

=

( )

!

!

n

n r

−

,

Thus

( )

!

P

!

n

r

n

n r

=

−

, where 0 < r

≤

n

This is a much more convenient expression for

n

P

r

than the previous one.

In particular, when r = n,

!

P !

0!

n

n

n

n

= =

Counting permutations is merely counting the number of ways in which some or

all objects at a time are rearranged. Arranging no object at all is the same as leaving

behind all the objects and we know that there is only one way of doing so. Thus, we

can have

n

P

0

= 1 =

! !

! ( 0)!

=

−

n n

n n

... (1)

Therefore, the formula (1) is applicable for r = 0 also.

Thus

( )

!

P 0

!

n

r

n

, r n

n r

= ≤ ≤

−

.

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