In how many ways can a committee of $3$ women and $4$ men be chosen from $8$ women and $7$ men if two particular women refuse to serve on the committee together?
I've approached this question by doing $^7C_4 ({}^7C_3+ {}^6C_3)$, since there are either $7$ or $6$ combinations when two of the women refuse to sit with each other?
Any help would be appreciated.
Best Answer
Not quite right. What you've done is find the number of ways of choosing 4 men from 7, and multiplied it by the number of ways of choosing the women. This is right so far.
Suppose the two special women who can't stand each other are Anna and Beth. Then either Anna, or Beth, or neither (but not both) can be on the committee. If Anna is, then you can choose $2$ of the remaining $6$ (we don't count Beth). Similarly for Beth. But if neither, then we can choose $3$ of the remaining $6$.