Combinatorics/Probability, Choosing from group of People 
I attempted to do this problem and I do have some guesses and trying to see whether they are right. Can you please correct if I'm wrong and explain. would really appreciate it. 
For a)
I have C(8,4) *C(9,2). So, 8 choose 4 * 9 choose 2. 
For b)
I have C(17,4) - C(8,4) - (9,9) [total - men - different ways to choose women.]
 A: If it's a four person committee, then you choose two men given that two women are chosen. So $\binom{9}{2} * \binom{8}{2}$ is your answer for (a).
For (b), break it up into cases. (a) gives you the case where there are two women. There are $\binom{9}{4}$ ways to choose a committee of four women. For three women, we have $\binom{9}{3} * \binom{8}{1}$. Each case is disjoint, so you add them up:
$\binom{9}{4} + \binom{9}{3} * \binom{8}{1} + \binom{9}{2} * \binom{8}{2}$.
A: a. The answer should be: $8 \choose 2$$\times$ $9 \choose 2$.
b. The answer is: $8 \choose 2$$\times$$9 \choose 2$ + $8 \choose 1$$\times$$9 \choose 3$ + $9 \choose 4$.
A: a)
$\binom{8}{2}*\binom{9}{2}$
I guess C(8,5) is a typo.
b)
$\binom{17}{4}-\binom{8}{4}-\binom{9}{1}*\binom{8}{3}$
total - all men - 1 woman 3 men 
A: I'm late to the party but why isn't a) You need to pick 2 female members: $\binom92$ and for the $2$ renaming positions we can choose between $8$ and $9-2=7$ female members (nothing specified that we need to pick men) so we have for a) $\binom{9}{2} * \binom{8 + 7}{2} = \binom{9}{2} * \binom{15}{2} = 3780$
and for question b) it is either 2 female and 2 rest or 3 females and 1 rest and 4 females and 0 rest so : $\binom{9}{2} * \binom{15}{2} + \binom{9}{3} * \binom{14}{1} + \binom{9}{4} * \binom{13}{0} = 5082$
