If two members refuse to be on the same team, how many teams are possible?

An engineering group consists of 12 men and 13 women.

If 2 women refuse to be on the same team together, how many different project teams can be formed consisting of 5 men and 5 women?

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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 $B$.

1. Count the number of ways to choose $5$ women from the $11$ more sociable ones, so that neither $A$ nor $B$ is chosen.
2. Count the number of ways to choose $A$ and $4$ of the $11$ more sociable women.
3. 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.

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12c5 = 792. 11c5 = 462. (11c4)*2 = 660 And now multiply all these? –  Imray Sep 24 '12 at 22:28
@Imray: No, you want to multiply the number of ways of choosing the men by the number of ways of choosing the women. When you choose the women, you don’t choose $5$ of the sociable ones and $5$ with $A$ but not $B$ and $5$ with $B$ but not $A$: you do one of those $\binom{11}5+2\binom{11}4$ things. So what two things should you multiply together? –  Brian M. Scott Sep 24 '12 at 22:31
Oh, so 2(11 4) is 11c4 for each unsociable. The total combinations in which one of them is in it is 2(11 c 4). So now I multiply (12 c 5) * ((11 c 5) + 2(11 c 4))? –  Imray Sep 24 '12 at 22:34
@Imray: The number of ways of choosing an acceptable group of $5$ women that includes $A$ plus the number of ways of choosing one that includes $B$. –  Brian M. Scott Sep 24 '12 at 22:35
so 2(11 4) is 11c4 for each unsociable. The total combinations in which one of them is in it is 2(11 c 4). So now I multiply (12 c 5) * ((11 c 5) + 2(11 c 4))? –  Imray Sep 24 '12 at 22:39

Hint: for the women, you can select 5 of the 11 who will work with anybody, or select 4 of 11 plus 1 of 2.

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Can you explain that? –  Imray Sep 24 '12 at 22:30