Number of ways a committee can be formed

In how many ways can a committee of four be formed from 10 men (including Richard)
and 12 women (including Isabel and Kathleen) if it is to have two men and two women

a) Isabel refuses to serve with Richard,

b) Isabel will serve only if Kathleen does, too

My Thoughts :
a) Total number of ways to select 4 people = 22C4 now for part a) I just need to deduct the
pairs where Isabel and Richard are both in the committe which = 20C2
I am not sure how to proceed with part (b) and if part (a) is entirely correct

• math.stackexchange.com/search?q=committee there are 1358 posts mentioning the word on Math.SE, did you read them all and unequivocally concluded that absolutely none of them was helpful? – user3733558 Apr 17 at 20:45
• $\binom{22}4$ counts a lot more than you want. You can get a committee that is all men or all women, for example. Start with $\binom{10}2\binom{12}2$ and then start subtracting. – Thomas Andrews Apr 17 at 20:54

In how many ways can a committee of four be formed from $$10$$ men (including Richard) and $$12$$ women (including Isabel and Kathleen) if it is to have two men and two women and Isabel refuses to serve with Richard?

We must subtract those committees on which both Isabel and Richard serve from the total number of committees formed with two men and two women. The number of committees that can be formed with two men and two women is $$\binom{10}{2}\binom{12}{2}$$ since we must select two of the ten men and two of the twelve women to serve on the committee. The number of committees with two men and two women which could be formed if both Richard and Isabel were to serve is $$\binom{1}{1}\binom{9}{1}\binom{1}{1}\binom{11}{1} = \binom{9}{1}\binom{11}{1}$$ since we would have to select one of the other nine men and one of the other eleven women to serve with Richard and Isabel. Therefore, the number of admissible committees is $$\binom{10}{2}\binom{12}{2} - \binom{9}{1}\binom{11}{1}$$

In how many ways can a committee of four be formed from $$10$$ men (including Richard) and $$12$$ women (including Isabel and Kathleen) if Isabel will serve only if Kathleen does, too?

If Isabel will only serve if Kathleen does, too, then there are two possibilities: Either they both serve or neither serves. If both Isabel and Kathleen serve, they must be the two women on the committee. If neither serves, then select two of the other ten women to serve on the committee. In either case, select which two of the ten men serve on the committee.

There are $$\left[\binom{2}{2} + \binom{10}{2}\right]\binom{10}{2}$$ admissible committees.

You could divide the problem into all the committees without Isabel and all those with her. $${10 \choose 2} {11 \choose 2}$$ for those without her. And $${9 \choose 2} * 1$$ for those with her.

Part (b). There are $$^{20}C_3$$ ways when Isabel is on and Kathleen is not.

Part (a) is correct.

• You overlooked the requirement that two men and two women must serve on the committee. – N. F. Taussig Apr 18 at 2:17
• Sorry. I missed that. – herb steinberg Apr 18 at 3:14