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Question

A dance class consists of $22$ students, of which $10$ are women and $12$ are men. If $5$ men and $5$ women are to be chosen and then paired off, how many results are possible?

Approach

According to me, the number of results possible is:

$$\binom{10}{5}*\binom{12}{5}*5!*2^{5}$$

Answer given :

$$\binom{10}{5}*\binom{12}{5}*5!$$

My conclusion

Shouldn't be there $2$ options in each pair i.e ordering between men and women for $5$ such group, making it $5!$? Why is the answer not leaving $5!$? Are they not considering order? And if the order is important, is my answer correct in this case?

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    $\begingroup$ Why would the be two options in each pair . A pair is of combination of a man and a woman . Even if you select fist a man then a woman to pair him or select first a woman and then a man to pair her both are same $\endgroup$ – ATHARVA Jun 1 '17 at 13:04
  • $\begingroup$ if order is important, ur answer is right i believe $\endgroup$ – Kiran Jun 1 '17 at 13:11
  • $\begingroup$ Yes, your answer is correct if "Fred and Ginger" is counted as a distinct pairing from "Ginger and Fred". $\endgroup$ – Michael Seifert Jun 1 '17 at 13:49
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$$\binom{10}{5}*\binom{12}{5}*5!*2^{5}$$ is the right answer if the order in which you pick the pair is important. For example, if(x,y) and (y,x) are different.

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First select five men and women. Then select one man $M_1$ at random, and pair him off with one of the 5 different women. Next select another man $M_2$, and assign him to one of the 4 remaining women. Continuing this, there are $5!$ ways to pair off the 5 selected men and women. As such, the total number of pairs equals:

$${10 \choose 5}{12 \choose 5}5!$$

The order in which the pairs are chosen does not matter.

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