# Selecting a committe with two women refusing to sit together.

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.

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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 $6$.

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Suppose Woman X and Woman Y refuse to sit each other. Then there are three cases to consider when choosing 3 of the 8 women.

Case 1: Suppose we select Woman X but not Woman Y. After selecting Woman X, there are $3-1=2$ more women to select from the other $8-2=6$ candidate women, so we have ${}_6C_2$ ways.

Case 2: Suppose we select Woman Y but not Woman X. After selecting Woman Y, there are $3-1=2$ more women to select from the other $8-2=6$ candidate women, so we have ${}_6C_2$ ways.

Case 2: Suppose we select neither Woman X nor Woman Y. Then we must select $3$ women from the other $8-2=6$ candidate women, so we have ${}_6C_3$ ways.

Hence, our final answer is: $${}_7C_4 \cdot ({}_6C_2 + {}_6C_2 + {}_6C_3)$$

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You’ve miscounted the ways to choose $3$ women willing to serve together.

There are $\binom83$ ways to choose $3$ women without any restriction. If we choose the two difficult women, there are $6$ ways to choose the third woman, so there are only $6$ unusable sets of $3$ women. Thus, the number of acceptable sets of $3$ women is $\binom83-6$, and the number of committees is

$$\binom74\left(\binom83-6\right)=35(56-6)=1750\;.$$

It’s possible to do it by counting the usable sets of $3$ women directly, but it’s a little harder. There are $\binom63$ ways to choose a group that has neither of the difficult women. To choose a group that has exactly one of them, we first choose which of the two we’ll include, and then we choose $2$ of the $6$ women who don’t care with whom they serve; this can be done in $2\cdot\binom62$ ways. We then get a total of

$$\binom74\left(\binom63+2\binom62\right)=35\cdot(20+2\cdot15)=1750$$

ways to form the committee.

To see exactly how you’ve miscounted, notice that your $\binom73$ corresponds to my $2\binom62$. You’ve picked one of the difficult women, set her aside, and chosen any $3$ from the remaining $7$. The problem here is two-fold. First, this count includes every group of $3$ women that does not include either of the difficult women, and you already counted those groups in your $\binom63$. Secondly, it doesn’t take into account the fact that there are $2$ ways to decide which difficult woman is not on the committee.

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