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