how many ways are possible...give a brief working ?

1) 5 men and 0 women => (6C5)*(4C0) = 6*1 = 6 ways

2) 4 men and 1 woman => (6C4)*(4C1) = 15*4 = 60 ways

3) 3 men and 2 women => (6C3)*(4C2) = 20*6 = 120 ways

so total 6+60+120 = 186 ways ............ answer

We first select 3, 4 or 5 men out of 6 men. We can't select less than 3 because there should be more men in committee than women.

For each of these selections, we need to select 0, 1 or 2 women out of 4 women total.

6 C 3 = 6! / ( 3! * 3! ) = 720 / (6*6) = 20. You can use Google "6 choose 3" to get this same answer.

6 C 4 = 15;

6 C 5 = 6;

4 C 0 = 1;

4 C 1 = 4;

4 C 2 = 6;

So there are 20 * 15 * 6 ways to choose men members = 1800 ways. And there are 1 * 4 * 6 ways to choose female members of the committee = 24 ways. In total, there are 1800 * 24 = 43200 ways to choose the whole committee.

@Additional details:

Oops. There are 20 + 15 + 6 = 41 ways to choose men members. And 1 + 4 + 6 = 11 ways to choose females. Total = 41 * 11 = 451 ways. The other people are not multiplying independent events of choosing males and choosing females.

First, consider the different ways in which a 5 member committee with more men can be formed.

Case 1 (5 men and 0 women)

- 6C5 = 6 ways

Case 2 (4 men and 1 woman)

- (6C4)*(4C1) = 60 ways

Case 3 (3 men and 2 women

- 6C3)*(4C2) = 120 ways

In this case, since the 3 cases are independent of each other, you add the number of ways for each case to get the total number.

total 6+60+120 = 186 ways

I had that same question in school and the teacher answerd this:

1) 5 men and 0 women => (6C5)*(4C0) = 6*1 = 6 ways

2) 4 men and 1 woman => (6C4)*(4C1) = 15*4 = 60 ways

3) 3 men and 2 women => (6C3)*(4C2) = 20*6 = 120 ways

so total 6+60+120 = 186 ways ............ answer

so there you go it's very easy to calculate this just by doing mental questions