There are
[tex]\dbinom{2n}2 = \dfrac{(2n)!}{2! (2n-2)!}[/tex]
ways of pairing up any 2 members from the pool of [tex]2n[/tex] contestants. Note that
[tex](2n)! = 1\times2\times3\times4\times\cdots\times(2n-2)\times(2n-1)\times(2n) = (2n-2)! \times(2n-1) \times(2n)[/tex]
so that
[tex]\dbinom{2n}2 = \dfrac{(2n)\times(2n-1)\times(2n-2)!}{2! (2n-2)!} = \boxed{n(2n-1)}[/tex]
