Aptitude - Permutation and Combination - Discussion

We need a minimum 3 men, so the maximum can be all 5 men in the committee.

3 men AND 2 women = 7C3 X 6C2 = 525.
4 men AND 1 women =7C4 X 6C1 = 210.
all 5 men = 7C5 = 21.

Add them 525 + 210 + 21 = 756.

How it's changed from 7c4 to 7c3? Please explain me.

Men=7, Women = 6.

Method -1 (correct).

First of all, we need to select 3 men from 7 so = 7C3 = 35.
In problem, they are asking for a minimum of 3 men so already you took 3 men.

So now we need women.
So we need 2 women from 6 women.
So 6c2.
So the value of 6c2 is 15.
Now 35*15 = 525.

In same way, the second committee we took 4 men from the 7 men group.

i.e 7c4 = 35.

We need one more woman for the 2nd committee because already we took 4 men.

So, 6c1 = 6.
So, 35*6 = 210.

Now we took all men for 3 rd committee.
So 7c5 = 21.

Now the total is 525+210+21 = 756.


METHOD II (Seems to be correct).

Why would the logic 7C3 * 10C2 not work? First, we've chosen 3 Men - 7C3. Now we have 10 people left (which included both men and women). Then we need to select any 2 people either men or women so 10C2.

Mistake: let 7 men be a, b, c, d, e, f, g.

7c3==> (a, b, c) ; (b, c, d) ; (c, d, e) ;. 35 sets.

Let us look into one condition.

4 men, 1women (let us assume that women is fixed).

(a b c) ;d;fixed women.
(b c d) a; fixed women.
(a b d);c;fixed women.

And so in a similar way, there will be the same items counted multiple times, in method 1 this does not occur.

Why can't it be, 7C3 (at least 3 men) * 10C2 (any 2 may be men or women or both out of the remaining 10)?

@All.

Here, some of the persons are confused in the second step because there was an incorrection where (7C4 x 6C1) + (7C5) were converted into + (7C3 x 6C1) + (7C2).

Anyone, Please explain this 7C3 * 6C1+7C2.

Thanks everyone for explaining the answer.

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

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.

3 men & 2 woman = 7C3 * 5C2 = 525.
4 men & 1 woman = 7C4 * 5C1 = 210.
5 men = 7C5 = 21.
So, the answer is 756.