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?
[Math] If two members refuse to be on the same team, how many teams are possible
combinatorics
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Best Answer
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 $B$.
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.