Hi,

A tennis tournament has a field of 2n entrants, all of whom need to be scheduled to play in the first round. How many different pairings are possible?

The solution is $\displaystyle {(2n)!\over{n!(2!)^n}} = 1 \cdot 3 \cdot 5 \cdot . . . \cdot (2n-1) $

But I cant see my way through the logic of this solution.

Thanks for any insights

MD