$12$ women and $10$ men are on the faculty. How many ways are there to pick a
committee of $7$ if
(a) Claire and Bob will not serve together,
(b) at least one woman must be chosen
I'm not sure how to start a. Essentially I need to remove a woman and remove a man?
For b, $\binom{22}{7}- \binom{10}{7}= 170424$ ways
Best Answer
(a) The total number of committees are $\binom{22}{7}$. If Claire and Bob serve together, we have two people fixed, and so we have $\binom{20}{5}$ such committees. And so if Claire and Bob will not serve together, we have $\binom{22}{7} - \binom{20}{5}$.
For (b), we use inclusion-exclusion. There are $\binom{22}{7}$ possible committees and $\binom{10}{7}$ committees with no women. We subtract the second from the first to get the final result.