# In how many ways a committee consisting of 5 men and 3 women, can be chosen from 9 men and 12 women?

• Last Updated : 20 Jan, 2022

A combination is a selection of items from a collection, such that the order of selection does not matter. Basically, it can be considered as an unordered permutation. A k-combination can be referred to as the combination of n things taken k at a time without repetition. For instance, we have, three alphabets A, B, and C, and we need to pick any two alphabets from it, the possible combinations are obtained: AB, BC, and AC. In the case of combinations, the order of selection of items doesn’t matter, that is AB is equivalent in nature to BA.

nCr = n!/(n – r)!r!

Where,

n – total number of objects from which to choose data

r – the number of chosen objects

### In how many ways a committee consisting of 5 men and 3 women, can be chosen from 9 men and 12 women

Solution:

Since, combinations are referred to using the following formula:

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

Since we need to choose, 5 men out of 9 men,

For Men = 9C5

Since, we need to choose, 3 women out of 12 women,

For Women = 12C3

Now,

Choose 5 men out of 9 men

Therefore,

nCr = n!/(n – r)!r!

9C5 = 9!/(9 – 5)!5!

9C5 = 9!/4!5!

9C5 = 9 × 8 × 7 × 6 × 5!/4 × 3 × 2 × 1 × 5!

9C5 = 3024 × 5!/24 × 5!

Simplifying Further,

9C5 = 3024/24

9C5 = 126

Thus,

We can choose 5 men out of 9 men in 126 ways

Now, further calculating for women,

Choose 3 women out of 12 women

Therefore,

nCr = n!/(n – r)!r!

12C3 = 12!/(12 – 3)!3!

12C3 = 12!/9!3!

12C3 = 12 × 11 × 10 × 9!/9! × 3 × 2 × 1

12C3 = 1320 × 9!/6 × 9!

Simplifying Further

12C3 = 1320/6

12C3 = 220

Thus,

Choose 3 women out of 12 women in 220 ways

Therefore,

The committee can be chosen in 126 × 220 = 27720 ways.

### Sample Problems

Question 1: illustrate some of the examples of permutation and combination?

Solution:

Question 2: Find the permutation and combination for n = 4 and r = 2?

Solution:

Permutation:

nPr = n!/(n – r)!

4P2 = 4!/(4 – 2)!

4P2 = 4!/2!

4P2 = 4 × 3 × 2!/2!

Simplifying

4P2 = 4 × 3

4P2 = 12

Combination:

nCr = n!/(n – r)!r!

4C2 = 4!/(4 – 2)!2!

4C2 = 4!/2!2!

4C2 = 4 × 3 × 2!/2 × 1 × 2!

Simplifying

4C2 = 4 × 3/ 2

4C2 = 12/2

4C2 = 6

Question 3: In how many ways flowers for a bouquet having 6 roses and 8 lilies, can be chosen from 8 roses and 10 lilies?

Solution:

Here choose flowers for bouquet

Finding how rose can be chosen

6 rose from 8 rose = 8C6 = 8!/(8-6)!6!

= 8!/2!6!

= 8 × 7 × 6!/2 × 1 × 6!

Simplifying

= 56/2

= 28

8C6 = 28

Roses can be chosen in 28 ways

Finding how lily can be chosen

8 lily from 10 lily = 10C8 = 10!/(10 – 8)!8!

= 10!/2!8!

= 10 × 9 × 8!/2 × 1 × 8!

Simplifying

= 90/2

= 45

10C8 = 45

Lily can be chosen in 45 ways

The flowers for bouquet can be chosen in 1260 ways.

Question 4: Evaluate C(22, 20). C(n, r) = n!/(n – r)!r!

Solution:

Here,

n = 22

r = 20

Putting the values of n and r in C(n, r) = n!/(n – r)!r!

C(22, 20) = 22!/(22 – 20)!20!

C(22, 20) = 22!/(2)!20!

C(22, 20) = 22 × 21 × 20!/2! × 20!

C(22, 20) = 22 × 21/ 2 × 1

C(22, 20) = 462/2

C(22, 20) = 231

Therefore,

C(22, 20). C(n,r)=n!/(n-r)!r! = 231

Question 5: Find out C(30, 28). C(n, r) = n!/(n – r)!r!

Solution:

Here,

n = 30

r = 28

Putting the values of n and r in C(n, r) = n!/(n – r)!r!

We get,

C(30, 28) = 30!/(30 – 28)!28!

C(30, 28) = 30!/(2)!28!

C(30, 28) = 30 × 29 × 28!/2! × 28!

C(30, 28) = 30 × 29/ 2

C(30, 28) = 870/2

C(30, 28) = 435

Therefore,

C(30, 28). C(n,r)=n!/(n – r)!r! = 435

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