# Aptitude - Permutation and Combination - Discussion

Discussion Forum : Permutation and Combination - General Questions (Q.No. 1)
1.
From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
564
645
735
756
None of these
Explanation:

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).

Required number of ways = (7C3 x 6C2) + (7C4 x 6C1) + (7C5)
 = 7 x 6 x 5 x 6 x 5 + (7C3 x 6C1) + (7C2) 3 x 2 x 1 2 x 1
 = 525 + 7 x 6 x 5 x 6 + 7 x 6 3 x 2 x 1 2 x 1
= (525 + 210 + 21)
= 756.

Discussion:
131 comments Page 1 of 14.

Sumanth reddy said:   6 months ago
How it's changed from 7c4 to 7c3? Please explain me.
(8)

Suga said:   2 years ago
@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).
(7)

Abi said:   2 years ago
Anyone, Please explain this 7C3 * 6C1+7C2.
(7)

Y Krishna Pavan Sai said:   2 years ago
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.
(10)

Anu said:   2 years ago
(3)

Johar said:   2 years ago
Anyone, Please solve this problem clearly to get it.
(1)

Jay said:   2 years ago
@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.
(2)

Shaifali said:   3 years ago
Well explained @Nagu.

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

Sake said:   3 years ago
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.
(2)

Akshay said:   4 years ago

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