# Combinations (1 Viewer)

#### CapitalSwine

##### New Member
I'm stuck on this question:

'In how many ways can a committee of 3 women and 4 girls be chosen from 7 women and 6 girls so that if the eldest woman is serving on the committee then the youngest girl is not?'

#### integral95

##### Well-Known Member
So you first consider the case where the eldest women is chosen, that means there's 2 other women to choose from 6 and 4 girls from 5 since one of them is out.

$\binom{6}{2}\binom{5}{4} = 75$

Then you consider the other case where the eldest women isn't chosen, that means there's 3 women from a group of 6 to choose from and a group of 4 girls from 6.

$\binom{6}{3}\binom{6}{4} = 300$

Add them both to get 375, is that the answer?

#### Roy G Biv

##### Member
alternatively, divide into 3 cases:
oldest woman in, youngest girl not
oldest woman out, younger girl in
oldest woman out, youngest girl out

you get the same answer
just understand what you're doing and think your way through. integral's way is 100% correct too.