# Permutation and Combination - Important Concepts and Notes

By Abhinav Gupta|Updated : August 21st, 2021

Permutation and combination are the ways to represent a group of objects by selecting them in a set and forming subsets. It defines the various ways to arrange a certain group of data. When we select the data or objects from a certain group, it is said to be permutations, whereas the order in which they are represented is called combination. Both concepts are very important in Mathematics.

Permutation and combination are the ways to represent a group of objects by selecting them in a set and forming subsets. It defines the various ways to arrange a certain group of data. When we select the data or objects from a certain group, it is said to be permutations, whereas the order in which they are represented is called combination. Both concepts are very important in Mathematics.

## Permutation and Combination: Notes & Important Questions

Permutation and combination

Permutations and combinations are the basic ways of counting from a given set, generally without replacement, to form subsets.

Permutation

Permutation of an object means the arrangement of the object in some sequence or order.

Theorem 1

The total number of permutations of a set of n objects taken r at a time is given by

P( n,r)  =  n (n- 1)(n- 2)….….(n- r+1)     (n ≥ r)

Proof:

The number of permutations of a set of `n’objects taken r at a time is equivalent to the number of ways in which r positions can be filled up by those n objects . When first object is filled then we have (n-1) choices to fill up the second position. Similarly, there are (n- 2) choices fill up the third position and so on.

Therefore, P(n,r)  =  n (n- 1) (n- 2) …….. (n – r +1)

=n(n−1)(n−2)…….(n−r+1)(n−r)…3.2.1(n−r)……3.2.1n(n−1)(n−2)…….(n−r+1)(n−r)…3.2.1(n−r)……3.2.1

=  n!(n−r)!n!(n−r)!

Permutation of objects not all different

The permutation of objects taken all at a time when P of the objects are of the first kind, q of them are of the second kind, of them are of the third kind and the rest all are different.

The total number of permutations =n!p!q!r!n!p!q!r!

Circular permutation

Circular Permutation: The number of ways to arrange distinct objects along a fixed line

The total number of permutation of a set of n objects arranged in a circle is P = (n -1)!

Permutation of repeated things

The permutation of the n objects taken r at a time when each occurs a number of times and it is given  by  P = nr

Example 1

How many numbers of three digits can be formed from the integers 2,3, 4,5,6? How many of them will be divisible by 5?

Soln:

For the three digits numbers, there are 5 ways to fill in the 1st place, there are 4 ways to fill in the 2nd place and there are 3 ways to fill in the 3rd place. By the basic principle of counting, number of three digits numbers = 5 * 4 * 3 = 60.

Again, for three-digit numbers which are divisible by 5, the number in the unit place must be 5. So, the unit place can be filled up in 1 way. After filling up the unit place 4 numbers are left. Ten’s place can be filled up in 4 ways and hundredths place can be filled up in 3 ways. Then by the basic principle of counting, no.of 3 digits numbers which are divisible by 5 = 1 * 4 * 3 = 12.

Example 2

How many numbers of at least three different digits can be formed the integers 1, 2, 34, 5, 6,?

Soln

Numbers formed should be of at least 3 digits means they may be of 3 digits, 4 digits, 5 digits or 6 digits.

There are 6 choices for the digit in the units place. There are 5 and 4 choices for digits in ten and hundred’s place respectively.

So, the total number of ways by which 3 digits numbers can be formed = 6.5.4 = 120

Similarly, the total no.of ways by which 4 digits numbers can be formed = 6.5.4.3 = 360.

the total no. of ways by which 5 digits numbers can be formed = 6.5.4.3.2 = 720.

The total no.of ways by which 4 digits numbers can be formed = 6.5.4.3.2.1 = 720.

So, total no.of ways by which the numbers of at least 3 digits can be formed = 120 + 360 + 720 + 720 = 1920

Example 3

In how many ways can four boys and three girls be seated in a row containing seven seats

1. if they may sit anywhere
2. if the boys and girls must alternate
3. if all three girls are together?

a.

Soln:

If the boys and girls may sit anywhere, then there are 7 persons and 7 seats. 7 persons in 7 seats can be arranged in P(7,7) ways.

= 7!(7−7)!7!(7−7)! = 7!0!7!0! = 7∗6∗5∗4∗3∗2∗117∗6∗5∗4∗3∗2∗11 = 5,040 ways.

b.

Soln:

If the boys and girls must sit alternately, there are 4 seats for boys and 3 for girls.

Here, for boys n = 4, r = 4

4 boys in 4 seats can be arranged in P(4,4) ways

= 4!(4−4)!4!(4−4)! = 4∗3∗2∗114∗3∗2∗11 = 24 ways.

Again. For girls n = 3, r = 3

3 girls in 3 seats can be arranged in P(n,r) i.e. P(3,3) ways.

= 3!0!3!0! = 3∗2∗113∗2∗11 = 6ways.

So, total no.of arrangement = 24 * 6 = 144 ways.

c.

Suppose 3 girls = 1 object, then total number of student (n) = 4 + 1= 5.

Then the permutation of 5 objects taken 5 at a time.

= P(5,5) = 5!(5−5)!5!(5−5)! = 5∗4∗3∗2∗115∗4∗3∗2∗11 = 120.

We know, 3 girls can be arranged themselves in P(3,3) different ways,

i.e. P(3,3) = 3!(3−3)!3!(3−3)! = 3 * 2 * 1 = 6 different ways.

Therefore, required of arrangements = 120 * 6 = 720.

Examples 4

In how many ways can eight people be seated in a round table if two people insisting sitting next to each other?

a.

Total number of objects = 4 + 4 = 8.

If they may sit anywhere, then it is the circular arrangements of 8 objects taken 8 at a time. So, the total number of permutation.

= (8 – 1)! = 7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040.

b.

Art and Science students have to sit alternately in a round table. So, there are 4 seats for Art students and 4 for Science students.

4 art students at a round table can be arranged in 4 – 1! ways.

= 3! = 3 * 2 * 1 = 6 ways.

Again. 4 science students can be arranged in P(4,4) ways, i.e. 4! Ways = 4 * 3 * 2 * 1 = 24 ways.

So, total no.of arrangement = 6 * 24 = 144 ways.

Combinations

The combination means a collection of an object without regarding the order of arrangement. The total number of combinations of n objects taken r at a time C(n,r ) is given by C(n,r) =  n!(n−r)!r!n!(n−r)!r!

Examples 5

From 4 mathematician, 6 statisticians and 5 economists, how many committees consisting of 3 men and 2 women are possible?

Soln:

2 members can be selected from 4 mathematicians in C(4,2) in different ways.

2 members can be selected from 6 statisticians in C(6,2) in different ways.

2 members can be selected from 5 economics in C(5,2) in different ways.

Therefore, total number of committees = C(4,2) * C(6,2) * C(5,2) = 6 * 15 * 10 = 900.

Examples 6

1. If C(20, r+ 5) = C (20, 2r  -7) find  C( 15,r)
2. if  C(n, 10)  +  C(n,9)  =  C( 20,10)  find n and C(n, 17)
3. solve for  n the equation C(n+ 2,4) = 6 C(n, 2)

a.

Given C(20,r + 5) = C(20,2r – 7)

Then r + 5 = 2r – 7      [if C(n,r) = C(n,r’) then r = r’]

Or, r = 12.

So, C(15,r) = C(15,12) = 15!(15−12)!.12!15!(15−12)!.12! = 455.

b.

Given. C(n,10) + C(n,9) = C(20,10)

Or, C(n + 1,10) = C(20,10)     [if C(n,r)+ C(n,r – 1) = C(n + 1, r)]

So, n + 1 = 20

So, n = 19.

Again, C(n,17) = C(19,17) = 19!(19−17)!.17!19!(19−17)!.17! = 171.

c.

C(n + 2,4) = 6 C(n,2)

Or, (n+2)!(n−2)!.4!(n+2)!(n−2)!.4! = 6. n!(n−2)!.2!n!(n−2)!.2!à(n+2).(n+1).n!(n−2)!.4!(n+2).(n+1).n!(n−2)!.4! = 6n!(n−2)!.26n!(n−2)!.2.

Or, (n+2)(n+1)4∗3∗2∗1(n+2)(n+1)4∗3∗2∗1 = 3

Or, n2 + 3n + 2 = 72

Or, n2 + 3n – 70 = 0

Or, (n + 10)(n – 7) = 0

Either, n = - 10 or, n = 7.

n = - 10 is not possible.

So, n = 7.

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