Aptitude - Permutation and Combination - Discussion

How 6C2 equal to 15 please explain?

In that question that is cleared that 5 members need to be select out of 7 men and 6 women.

And in question there is a condition that at least 3 men that mean we can select in 3 ways that we can select 3 men or 4 men or 5 men.

1) 3 men and we want another 5-3 = 2 women. We have to select 3 men from 7 men that is 7C3 and 2 womens from 6 women that is 6C2 results is 7C3*6C2.

2) 4men and we want another 5-4 = 1 women. We have to select 4 men from 7 men that is 7C4 and 1 women from 6 women that is 6C1 results is 7C4*6C1.

3) 5 men and we want another 5-5 = 0 women. We have to select 5 men from 7 men that is 7C5 and 0 women from 6 women that is 6C2 results is 7C5*6C0.

Answer is we need to sum up all these combinations that is (7C3*6C2+7C4*6C1+7C5*6C0) that is 756.

I am still have little problem with the solving.

I am still have little problem with the solving.

Why can't we select only five women out of 6 women. Why that combination is not taken into account.

The question is not specific regarding the selection of men and women after the selection of at least three men.

If we take the method [7C3 * 6C2] + [7C4 * 6C1] + [7C5 * 6C0] as the solution, then we must understand that all men are taken as one entity first then women are considered as another entity.

Hence, 3 men are one entity and 2 women are another entity. Similarly, 4 men are one and likewise, 5 men are one entity.

That means if you are done with the selection of 3 men at least, then DONT SELECT MEN AGAIN from the remaining. And if you choose to select men again, THEN INCLUDE THEM WHILE SELECTING ALTEAST THREE MEN.

I understood all the combination how we choose, Thanks for this @ Mohan & @Pavz. :)

But I have little doubt in solving cases like;

1) why we "multiple" in case?

2) why we "add" all cases?

@Malu.

The problem says at least 3 men so if you select 5 women it would contradiction to the problem.

Where do you get from 35?

Kindly, solve this.

The committee of six is to be formed from a group of seven engineers and four mathematicians, how many different committees can be formed if at least two mathematicians are always to be included?