0
$\begingroup$

I am trying to find an answer to the probability of finding a number of ways to take action.

Class that has 20 students. 11 women and 9 men.

Want to randomly select 6 men and 6 women.

After selection will make six pairs (each pair including one man and one woman). How many ways can I choose?

I tried the following: $$\binom{9}{6} \binom{11}{6}(6!)(6!) = 2.011 * 10 ^{10}$$

It seems to me a number too large for it to be sensible

$\endgroup$
  • 1
    $\begingroup$ There is one factor $6!$ too much. If $6$ men and $6$ women are chosen ($\binom96\binom{11}6$ ways for that) then there are $6!$ ways for matchmaking. Put the $6$ men in a row and start "distributing" the women. $\endgroup$ – drhab Mar 15 '16 at 15:22
  • $\begingroup$ Why? I Can switch between men and women each six times in groups ... $\endgroup$ – BAM Mar 15 '16 at 15:24
4
$\begingroup$

As drhab pointed out, you've got an excess factor of $6!$. You can choose $6$ men and $6$ women; then you can put the men in a row, and then the women have $6!$ choices of pairing up with the men. Permuting the positions of the men in the row doesn't increase that number of choices.

$\endgroup$

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.