COMBINATIONS

If you have not gone through Permutation, please go through it before proceeding with this article.

Combinations are the way one can select a thing from a set of things. In Combinations, the order in which the set is arranged does not matters.

For eg.

** Question 1 –** There are 5 people in audition left. 2 of them will be selected to host the show. How many ways this can be done?

In above questions, order does not matter. It does not matter who is selected first and who is selected second, what matters is that 2 people are selected.

Suppose the 5 people presents are A, B, C, D, and E.

The combinations of 2 people selected could be ->

- AB
- AC
- AD
- AE
- BC
- BD
- BE
- CD
- CE
- DE

There are 10 combinations that could be selected. Make sure to note that AB and BA are not different, they are same. However, if this has been an arrangement question we would have considered AB and BA as different entity. Hence Combinations can be defined as,

Combinations = Permutation / (Number of way to arrange things)

**Number of ways to arrange n things = n!**

Number of ways to select 2 people from 5 people

= (Number of ways to select 2 people from 5 people)/ (Number of way to arrange 2 people)

= (5! / (5-2)!) / 2!

= 5! / (3! * 2!)

= 5 * 4 * 3! / (3! * 2!)

= 5 * 4 / 2!

= 5 * 4 / (2 * 1)

= 5 * 2

= 10

**Number of ways to select r things in n things **

**= n!/((n-r)! * r!)**

**Question 2** – How many ways the 3 numbers can be selected from 2,5, and 8 with no repetition.

There are 3 numbers given and it is asked to select 3 numbers. In how many ways do you think this can be accomplished?

Did you said, 1 way. That’s right. There are only one way to select all three numbers.

Let’s calculate using the formulae,

**Number of ways to select r things in n things **

**= n!/((n-r)! * r!)**

Number of ways to select 3 things from 3 things

= 3! / (3-3)! * 3!

= 3! / (0! * 3!)

= 3! /(1 * 3!)

= 1

**Question 3 –** There are 5 Men and 4 Women presents out of which we need to select 2 Men and 2 Women. How many ways the selection can be done?

Total selection = (Number of ways to select 2 Men) * (Number of ways to select 2 Women)

Number of ways to select 2 Men from 5 Men = 5! / ((5-2)! * 2!)

= 5! / (3! * 2!)

= 5 * 4 * 3 *2 * 1 / (3* 2 * 1 * 2 * 1)

= 10

Number of ways to select 2 Women from 4 Women = 4! / ((4-2)! * 2!)

= 4! / (2! * 2!)

= 4 * 3 *2 * 1 / (2 * 1 * 2 * 1)

= 6

Total selection = (Number of ways to select 2 Men) * (Number of ways to select 2 Women)

= 10 * 6

= 60