How many ways can a committee of 3 men and 2 women be chosen from a pool of 8 men and 7 women?

Let $1,2,3,4,5,6,7,8$ be the women, and the men are denoted similarly, but with a hat, $\hat1$, $\hat2$, $\hat3$, $\hat4$, $\hat5$, $\hat6$.

The $1$ is not willing to work with $\hat 1$.

Now let us look at the counting strategy. Where do we count the configuration $2,3,4; \hat2,\hat3,\hat 4$?

• Well, at point one at any rate, it is one valid case among the many $\binom 73\binom 53$, since $1$ and $\hat 1$ are both excluded.
• Well, at point two also, it is one valid case among the many $\binom 83\binom 52$ cases, since $\hat 1$ is excluded.
• Well, at point three again, it is one valid case among the many $\binom 72\binom 63$ cases, since $1$ is excluded.

For a correct answer, count all possible commitees, there are $\binom 83\binom 63=1120$ of them, and subtract those where the pair $1,\hat 1$ is in the commitee, there are $\binom 72\binom 52=210$ of them. So the answer is $$ \binom 83\binom 63 - \binom 72\binom 52 =1120 -210 =910\ . $$

In the line "Probability that the first two selected are men with the third a woman," the probability of choosing a woman third, after choosing two men, is 7/13. You need to multiply by that as well. Similarly, the probability of choosing the woman first is missing from the following line. Finally, you also need to add in the additional case where a man is chosen first, a woman second, and another man third.

Another approach:

Number of committees consisting of two men and one woman: $\binom{8}{2}\binom{7}{1}$. Number of committees consisting of only men: $\binom{8}{3}$. The number of all possible committees is $\binom{15}{3}$. I think you can take it from there :)