# Math Help - Comb application 2

1. ## Comb application 2

in how many ways can a committee of five be selected from 8 women and 5 men.

a) without restriction
b) if there must be three women on the committee

2. Originally Posted by johnsy123
in how many ways can a committee of five be selected from 8 women and 5 men.

a) without restriction
b) if there must be three women on the committee
a) There are 13 people to choose from and you must choose 5.

So the number of combinations of 5 people from 13 is $\displaystyle {13\choose{5}} = \frac{13!}{5!(13 - 5)!}$.

For part b) does that mean there must be EXACTLY three women or AT LEAST three women?

3. never mind guys, i worked it out.

4. You're welcome :P

5. Your working out is wrong, because you can't simply work out the number of combinations as there are independent variables, so what i did is i worked out the combinations of which you can variate 4 men and 8 women into a committee of 5, there were 5.

-If i knew how to use the symbols on MHF, i would show you what i done.

6. Originally Posted by johnsy123
Your working out is wrong, because you can't simply work out the number of combinations as there are independent variables, so what i did is i worked out the combinations of which you can variate 4 men and 8 women into a committee of 5, there were 5.

-If i knew how to use the symbols on MHF, i would show you what i done.
But there are 5 men... And since you're choosing any 5 people out of those 13, it doesn't matter how many men and women there are. It's just the number of ways you can pick 5 people from 13, i.e. $\displaystyle {13\choose{5}}$.

7. Originally Posted by Prove It
But there are 5 men... And since you're choosing any 5 people out of those 13, it doesn't matter how many men and women there are. It's just the number of ways you can pick 5 people from 13, i.e. $\displaystyle {13\choose{5}}$.
I agree.

I assume the OP no longer needs (or no longer thinks s/he needs) help with part (b) ....?