[Math] Selecting 5 pairs of men and 5 women from 10 women and 12 men

combinatoricsdiscrete mathematics

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?

Best Answer

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