TITLE: Combination of splitting elements into pairs
QUESTION [23 upvotes]: Here is the problem:

Suppose there is a group of 12 people. In how many different ways can the 12 people be split into six pairs?

The answer is supposed to be $\frac{12!}{2^{6} 6!}$, but what I get is $\binom{12}{2} \binom{10}{2} \binom{8}{2} \binom{6}{2} \binom{4}{2} \binom{2}{2} = \frac{12!}{2^{6}}$.
Could anybody explain why is the factorial part necessary? I think it is related to the 6! possible permutations of the pairs but why is this relevant?
Thanks!

REPLY [10 votes]: This is an old question, but I think it could still benefit from a simpler way to derive the solution. There are 12! ways to arrange 12 people. Each permutation corresponds to a way of splitting them in pairs. In order not to overcount, we should account for the order in each pair (divide by two 6 times) and for the fact that we can freely permute the pairs themselves (divide by 6!). No binomial coefficients necessary.