Aptitude - Permutation and Combination - Discussion

I haven't understood that point. Anyone, please help me to get it.

Anyone, Please solve this problem clearly to get it.

@Nivetha.

We only have to find total combinations, not the probability. If we want to find probability then we have to divide it by 13C5.

Well explained @Nagu.

But I don't understand why do we need to multiply? why can't we add?

Why can't we do it like this,

Total ways = at least 3 men + not men.
i.e not men = 6c5.
Total ways= 13c5.
therefore,
13c5 - 6c5.

Yes it's write answer.

Using combination formula.
nCr = n! / r! ( n - r ) !
Question says at least 3 men's , means we have to select minimum 3 men's , it means we can select either 3 men's or 4 men's or 5 men's.
1)in first case select 3 men's and 2 women's. To form 5 people commette ..
7! / 3! (7-3)! * 6! / 2!(6-2)! - Eq-1
2 ) in second case select 4 men's and 1 women can be selected in.
7! / 4!(7-4)! * 6! / 1!(6-1)! - Eq -2
3) in third cas we can select 5 men's only
Can be selected in 7! / 5!(7-5)! - Eq -3

The final answer is:

Eq-1 + Eq-2 + Eq-3=
7! / 3! (7-3)! * 6! / 2!(6-2)! + 7! / 4!(7-4)! * 6! / 1!(6-1)! + 7! / 5!(7-5)!
= 525 + 210 + 21
=756 ways.

Shouldn't we have to divide the whole answer by 13c5? Please tell me.

There are 5members required so why don't we divide with 11c5?

Here 5 persons are select, and at least 3 men including this committee so we have
Men. Women
7c3 * 6c2 = 525
7c4 * 6c1 = 210
7c5 * 6c0 = 021
= 756 is the answer.

Why can't we do this? Select 3 out of 7 men. Because that's the main condition. Then select 2 women out of 6/1 man out of 4 and 1 woman out of 6/ 2 men out of 4.

i.e 7c3 (4c2 + 6c2 + 4c1. 6c1)?

Please tell me.