# Number of ways a committee could be formed if Mr $A$ refuses to serve on the committee if Ms $B$ is a member

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.

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).
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: $$\displaystyle \binom{9}{5}\cdot \binom{8}{6}=3528$$.
Subtract the bad committees (that contain both A and B) from the unrestricted ones: $$\binom{10}{ 5}\binom{8}{ 6}-\binom{9 }{ 4}\binom{7}{ 5}=4410.$$