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... how many ways can we make a team of 10 people consisting of 4 men and 6 women, such that 2 of the women are captains. My answer is $$\frac{1}{2!}\binom{25}{4}\frac{\binom{25}{6}}{N}$$ where $\binom{25}{4}$ is the number of ways we can pick 4 men, $\binom{25}{6}$ is the number of ways we can pick 6 women and the $2!$ is the overcounting factor. I know I have to divide $\binom{25}{6}$ by some other overcounting factor N, but I can't figure out what it is.

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    $\begingroup$ If there only two co-captains are to be named, it is $\binom{25}{4}\binom{25}{6}\binom{6}{2}$. There is no overcounting factor. Can you see where the $\binom{6}{2}$ comes from? $\endgroup$ – André Nicolas Mar 16 '16 at 0:04
  • $\begingroup$ Yes. $\binom{6}{2}$ is the number of ways we can pick 2 captains out of the 6 chosen women. Thanks. $\endgroup$ – Jeze Ken Mar 16 '16 at 0:48
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Community wiki answer to the question can be marked as answered:

Solution by André Nicolas in a comment:

Choose $4$ of $25$ men and $6$ of $25$ women, then choose $2$ of $6$ women as captains, for a product of

$$ \binom{25}4\binom{25}6\binom62=\frac{25!\cdot25!\cdot6!}{21!\cdot4!\cdot19!\cdot6!\cdot4!\cdot2!}=\frac{25!^2}{21!\cdot19!\cdot4!^2\cdot2!}\;. $$

Solution by Alex in an answer now deleted:

Choose $4$ of $25$ men, $2$ of $25$ women as captains and $4$ of the remaining $23$ women as non-captains, for a product of

$$ \binom{25}4\binom{25}2\binom{23}4=\frac{25!\cdot25!\cdot23!}{21!\cdot4!\cdot23!\cdot2!\cdot19!\cdot4!}=\frac{25!^2}{21!\cdot19!\cdot4!^2\cdot2!}\;. $$

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