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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

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    $\begingroup$ 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
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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.

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