Count the number of ways to choose $5$ men from the $12$ available; I think that you know how to do that. Now call the awkward women $A$ and $B$.
- Count the number of ways to choose $5$ women from the $11$ more sociable ones, so that neither $A$ nor $B$ is chosen.
- Count the number of ways to choose $A$ and $4$ of the $11$ more sociable women.
- Do the same for $B$.
Alternatively, you can count the number of unusable $5$-woman groups, i.e., those that contain both $A$ and $B$, and subtract that from the number of all possible $5$-woman groups.
Now you have the total number of ways in which you can choose the women. Multiply by the number of ways to choose the men, and you’re done.
First it might be helpful to remember that the binomial coefficients are given by
$$
{n \choose k} \;\; =\;\; \frac{n!}{k!(n-k)!}.
$$
Assuming that in your notation $P(13,7) = {13 \choose 7}$ then you're correct for the first part of the question. For the second part, you need to pick $4$ women and $3$ men. You have the right idea in what (I believe) you're writing, but it needs to be refined a bit.
Note that we need to pick $4$ out of seven women and there are ${7\choose 4}$ ways to do this. Similarly there are ${6\choose 3}$ ways to choose the men. Since the choice of women can be done independently of the choice of men (every potential group of women can be combined with any of the possible groups of men) we simply multiply these quantities together. The answer should be ${7\choose 4}{6\choose 3}$.
When we get to questions of there being "at most" or "at least" a certain number of men and/or women, we need to consider each case separately and then add all of the possible numbers together. To illustrate, consider the problem of computing the number of groups containing at least one man. We can make a list of all of the different possible numbers simply by considering each case:
- The number of groups with zero men is ${7 \choose 7}{6\choose 0} = 1$, since all $7$ women just get chosen.
- The number of groups with one man is ${7\choose 6}{6\choose 1} = 42$.
- The number of groups with two men is ${7\choose 5}{6\choose 2}$.
- etc.
To find the number of cases with at least one man, we add up the numbers of groups with one man, two men, three men, all the way to six men. Another way of computing this number too is to realize that the total number of possible groups is just ${13\choose 7}$. The only possible case that we are eliminating from our analysis is when all $7$ women get chosen, thus we can equivalently come to the conclusion that the number of groups with at least one man is ${13\choose 7} - {7\choose 7}{6\choose 0}$.
Best Answer
Let $X_1,X_2,\ldots,X_{13}$ the members of the team, suppose that $X_1$ and $X_2$ are the two persons that refuse to work together. Then, let's to count how many ways to form groups there exist.
If $X_1$ is in the group, then we must choose $6$ elements from the set of $11$ members $\{X_3,\ldots,X_{13}\}$. There are $\color{red}{{11 \choose 6}}$ ways to do this.
If $X_2$ is in the group, we can use the same reasoning as above, giving us $\color{red}{{11 \choose 6}}$ groups.
Finally, if $X_1$ and $X_2$ aren't in the group we have $\color{red}{{11 \choose 7}}$ different groups.