Chess tournament combinatorics

A much neater way is to note that if the players are arbitrarily ordered, the first player has $2n-1$ possible matchups, the second unmatched player has $2n-3$ and so on. Thus there are $(2n-1)(2n-3)\cdots(3)(1)$ possible matchups, which is usually denoted as $(2n-1)!!$ using the double factorial.

That will be $\frac{2n!}{2^{n}n!}$