... 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.

  • 1
    $\begingroup$ 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

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.