1. Combinations

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?'

2. Re: Combinations

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.

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

3. Re: Combinations

I second integral in this

4. Re: Combinations

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.

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