The term 'permutations and combinations' both have different meanings. Permutation is an arrangement of items, where its order matters. For instance, take any three names of your friends as follows: Martin (M), Victor (V) and Bob (B). |

$^nP_r$ = n (n - 1) (n - 2) .... (n - r + 1)

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

If repetition of object is allowed, then $^nP_r$ = n$^{r}$

### Solved Example

**Question:**Find the number of permutation, where value of ‘n’ is 7 and the value of ‘r’ is 4.

**Solution:**

**Permutation**$^nP_r$

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

Given, the value of n = 7 and value of r = 4.

First, we can find the factorial of 7.

The factorial of 7 is 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040.

Now, we find the factorial of (7 - 4).

The factorial of (7 - 4) is (7 - 4)! = 3!

So, the factorial of 3 is 3 x 2 x 1 = 6

Now, divide 5040 by 6

Permutation = $\frac{5040}{6}$ = 840.

**$^nP_r$**

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

Where, the values of ‘n’ and ‘r’ are the non- negative integers. Also, the value of ‘r’ is less than or equal to ‘n’.

‘r’ denotes the size of each permutation.

‘n’ denotes the size of the set from which element are permuted.

‘!’ denotes the factorial operator.

Each of the groups or selections that can be made by taking some or all of a number of things is called a combination. Thus, the selections that can be made by taking the letters a, b, c two at a time are ab, bc and ac. These three groups or selections are called the combination of three things taken 2 at a time. The number of combinations of n dissimilar things taken 'r' at a time is $^nC_r$.

In combination, order does not matters or we can say it is an unordered collection of a unique size. Now, we will define combination with the help of formulas.

Formula to find the combination is given as follows:

**$^nC_r$**

Combination =

Combination =

**=**$\frac{n!}{r! (n – r)!}$

Let us understand combination with the help of an example.

### Solved Example

**Question:**Suppose value of ‘n’ is 8 and value of ‘r’ is 5. Then, find the combination for the given value.

**Solution:**

**Combination =**$^nC_r$

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

Given, n = 8 and r = 5.

$^8C_5$ = $\frac{8!}{5! (8 - 5)!}$

Combination = 56.

$^nC_r$ = $\frac{n!}{r! (n – r)!}$

where, the values of ‘n’ and ‘r’ denote non-negative integers and value of ‘r’ is smaller than or equal to value of ‘n’. ‘r’ shows the size of each combination, ‘n’ shows the size of the set and ‘!’ denotes the factorial operator.

Given below are some of the example problems on combinations.

### Solved Examples

**Question 1:**Find the number of ways in which 5 players out of 8 players can be selected to form a team.

**Solution:**

**Question 2:**Out of 4 men and 3 women, a committee of 3 members is to be formed so that, it has 1 woman and 2 men. In how many different ways can this be done?

**Solution:**

By fundamental principle, these two can be done in ($^3C_1$ x $^4C_2$) = 18 ways.

### Solved Examples

**Question 1:**How many 4 digit numbers can be made from digits 1, 2, 3, 4, 5, 6 with no two digits common or repeated?

**Solution:**

$^6P_4$ = $\frac{6!}{(6-4)!}$

= $\frac{6!}{2!}$

= 360

Therefore, using four digit numbers, 360 numbers can be made with no two digits repeated.

**Question 2:**There are 4 candidates for the post of a lecturer in Mathematics and one is to be selected by votes of 5 men. Find the number of ways in which the votes can be given.

**Solution:**

5 men can vote in $4^{5}$ = 1024 ways.

Therefore, the votes can be given in 1024 ways.

### Solved Examples

**Question 1:**In how many ways can 3 people be seated in a row containing 6 seats?

**Solution:**

Then, by fundamental principle, total no of ways on which 3 person can be seated in 6 seats in a row is (6 x 5 x 4) ways = 120 ways.

**Question 2:**Three men have 3 coats, 4 waist coats and 5 caps. In how many ways can they wear them?

**Solution:**

Hence, by fundamental principle, the required number of ways = $^3P_3$ x $^4P_3$ x $^5P_3$ = 8640 ways.

**Question 3:**In how many different ways can the letters of the word "FLOWERS" be arranged?

**Solution:**

**Question 4:**In how many different ways can 2 boys and 4 girls be arranged on a row, such that all the boys stand together and all the girls stand together?

**Solution:**

The 4 girls are always together and can be arranged among themselves in $^4P_4$ ways = 4! ways.

The boys and girls can be arranged among themselves in $^2P_2$ ways = 2! ways.

So, total possible arrangements = 2! x 4! x 2! ways = 96 ways.