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In how many ways can a committee of $5$ women and $6$ men be chosen from $10$ women and $8$ men if Mr $A$ refuses to serve on the committee if Ms $B$ is a member.

What I tried:

We have two possibilities:

If Ms $B$ is a member of the committee, then ways $\displaystyle \binom{10}{4}\cdot \binom{7}{6}$

If Ms $B$ is not a member, then total ways $\displaystyle \binom{10}{5}\cdot \binom{8}{5}$

So total ways is addition of these two cases.

But answer given as $4410$. How do I solve it? Help me.

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You are counting some combinations multiple times. In the first case, Ms $B$ is already counted in, so there is $9$ women from which to choose. In the second case, as Ms $B$ is not included, there is only $9$ women from which to choose (and you should be choosing $6$ men).

Summing things up:

If Ms $B$ is a member and Mr $A$ not: $\displaystyle 1\cdot\binom{9}{4}\cdot \binom{7}{6}=882$

If Ms $B$ is not a member, but Mr. $A$ can be: $\displaystyle \binom{9}{5}\cdot \binom{8}{6}=3528$.

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Subtract the bad committees (that contain both A and B) from the unrestricted ones: $$\binom{10}{ 5}\binom{8}{ 6}-\binom{9 }{ 4}\binom{7}{ 5}=4410.$$

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