# Different ways of picking a committee of $12$ women and $10$ men

$12$ 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

• Hi there. For (a), try to count separately how many committees (i) include Claire, but not Bob, (ii) include Bob, but not Claire, (iii) include neither Claire, nor Bob. Apr 23, 2014 at 19:21

(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.

• I think your $b)$ is wrong, I asked the same question here math.stackexchange.com/questions/1828870/…
– Ovi
Jun 16, 2016 at 18:49
• Thanks for catching that! I've revised the answer. Jun 16, 2016 at 19:49
• "subtract the first from the second" - surely, we subtract the second from the first? Mar 4, 2018 at 16:26
• Thanks for catching that! Mar 4, 2018 at 19:03

For part a, I would use a sum: (without Bob and Claire) + (with only Bob) + (with only Claire) = all possibilities where both Bob and Claire are not together. For the groups formed with only one of them, say Bob, there's only 6 selections left to make up the committee of 7 so r = 6. Additionally, once we've decided Bob is on the committee, then the pool of remaining people that can be chosen is reduced by both himself (he's already selected) and Claire (don't want her) so the pool of people is now fewer by 2 people, n = 20.

Without Bob & Claire: 20 C 7 = 77520
With Bob only: 20 C 6 = 38760
With Claire only: 20 C 6 = 38760 Sum total = 155040