Question

A tennis tournament has 2n contestants. We want to pair them up for the first round of singles matches. Show that the number of different ways in which first round matches can be conducted is

Answer 1

There are

$\dbinom{2n}2 = \dfrac{(2n)!}{2! (2n-2)!}$

ways of pairing up any 2 members from the pool of $2n$ contestants. Note that

$(2n)! = 1\times2\times3\times4\times\cdots\times(2n-2)\times(2n-1)\times(2n) = (2n-2)! \times(2n-1) \times(2n)$

so that

$\dbinom{2n}2 = \dfrac{(2n)\times(2n-1)\times(2n-2)!}{2! (2n-2)!} = \boxed{n(2n-1)}$