 # Permutation and Combination

## Introduction

Permutation and Combination are the methods of counting which helps us to determine the number of different ways of arranging and selecting objects out of a given number of objects, without actually listing them. In this article, you will be able to learn the daily life application of permutation and combination along with their proper meaning and formula.

Before discussing permutation and combination, we will learn an important mathematical term ‘Factorial’.

### What is Factorial?

Factorial is defined as the product of all natural numbers less than or equal to a given natural number.

For a natural number ‘n’, its factorial is denoted by n!

And, n! = n (n-1) (n-2) (n-3) (n-4) …... 3 x 2 x 1.

For example: 7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040

10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 3628800

Note: (1) 0! = 1

(2) 1! = 1

### What is Permutation?

The arrangement of objects in a definite order is called Permutation.

For example: You have two books, one book of each subject, Mathematics and Science. You want to keep them properly on a shelf. So, you can keep them either Mathematics book first and then science book (i.e. M S), or Science book first and then Mathematics book next to it (i.e. S M). Therefore, there are two ways to arrange the two books on a shelf.

The number of permutations of n different objects taken r objects out of them without replacement, where 0 < r ≤ n, is given by:

nPr = $\frac{n!}{(n-r)!}$

From previous example, the number of permutations of 2 different books, Mathematics and Science, taken both of them is:

2P2 = $\frac{2!}{(2-2)!}$ = $\frac{2!}{0!}$ = 2! = 2 x 1 = 2 ways

### What is Combination?

The selection of some or all objects from a given set of different objects where order of selection is not considered is called Combination.

For example: Suppose you have three friends, Aman, Mohit and Raj, you want two of them to go with you for a picnic. You are not able to decide which two of them you should take for picnic, so you think of an idea that you will write each of their names on a separate paper and pick two of them. Then,             the possible ways you could get the names of your two friends will be:

1. Aman and Mohit

2. Mohit and Raj

3. Aman and Raj

Therefore, there are three ways to select two friends out of three friends.

The number of combinations of n different objects taken r objects out of them without replacement, where 0 < r ≤ n, is given by:

nCr = $\frac{n!}{r!(n-r)!}$ = $\frac{ₙPr}{r!}$

From previous example, the number of ways of picking the names of two friends out of three names is:

3C2 = $\frac{3!}{2!(3-2)!}$= $\frac{3!}{2! Х 1!}$ = $\frac{3Х2Х1}{2Х1}$ = 3 Ways

Note:

(1) If in a problem statement, you are asked for selection and their ordering, then you should use Permutation.

In simple words, Permutation = Selection + ordering

(2) If in a problem statement, you are asked only for selection then you should use Combination.

In simple words, Combination = Selection

## Difference Between Permutation and Combination

 Permutation Combination The number of ways to arrange objects The number of ways to select objects order is considered Order is not considered Clue words: arrangement, order, placed Clue words: select, group, choose It is denoted by nPr = $\frac{n!}{(n-r)!}$ It is denoted by nCr = $\frac{n!}{r!(n-r)!}$ = $\frac{ₙPr}{r!}$ Example: Arranging books, numbers, alphabets Example: selecting team members, clothes

### Solved Examples:

Q.1. How many words can be formed from the letters of the word ‘VEDANTU’ using each letter only once?

Solution: There are 7 letters in the word ‘VEDANTU’.

The number of ways of arranging all the 7 letters of given word is:

7P7 = $\frac{7!}{(7-7)!}$ = $\frac{7!}{0!}$ = 7! = 5040 words

Q.2. How many 5digit telephone numbers can be formed if each number starts with 21 and no digit appears more than once?

Solution: Since, the first two places have to be filled only by 2 and 1 respectively and there is only 1 way for doing this.

 2 1

Also, it is given that no digit appears more than once. Hence, there will be 8 digits remaining (0,3,4,5,6,7,8,9) and 3 places have to be filled with these remaining digits.
So, the next 3 places can be filled with the remaining 8 digits in 8P3 ways.

Total number of ways = 8P3 = $\frac{8!}{(8-3)!}$ = $\frac{8!}{5!}$ = $\frac{8Х7Х6Х5Х4Х3Х2Х1}{5Х4Х3Х2Х1}$ = 8×7×6 = 336

Q.3. How many chords can be drawn through 15 points on a circle?

Solution: Since, a chord is drawn by joining two points on a circle.

Given that, there are 15 points on a circle. The total number of ways of selecting two points on a circle will give the total number of chords of the circle.

Therefore, the required number of chords = 15C2 = $\frac{15!}{2!(15-2)!}$ = $\frac{15!}{2!(13)!}$ = $\frac{15Х14}{2Х1}$ = 105