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 $\displaystyle \displaystyle {13\choose{5}} = \frac{13!}{5!(13 - 5)!}$.
For part b) does that mean there must be EXACTLY three women or AT LEAST three women?
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.