TITLE: Basic combinatorics question
QUESTION [5 upvotes]: In a tennis tournament there are $2n$ participants. In the first round of the tournament,
each player plays exactly once, so there are $n$ games. Show that the pairings for the first
round can be arranged in exactly $(2n -1)!/2n^{n-1}(n-1)!$ ways.
I wondering if anyone could come up sith an elegant way, rather than a complicated argument with a lot of manipulation.
I'm 15, so I would appreciate if anyone could help me. Thanks.

REPLY [6 votes]: Answer:
The first pair could be chosen in ${2n\choose2}$.  Having chosen the first pair, the second pair could be chosen in ${2n-2\choose2}$.  Keeps going until you chose ${2\choose2}$ for the nth pair.  Thus the number of ways you could have pairings is ${2n\choose2}.{2n-2\choose2}.\cdots{2\choose2}$.  This you divide by n! ways to get the total number of ways 2n players could be chosen to play n games to eliminate ordering. Thus the answer after you breakdown each is
$$\dfrac{(2n)!}{2^n. n!}$$
This you could further complicate (just joking) to your answer by splitting the factorials such as below:
$$ \dfrac {2n.(2n-1)!}{2.2^{n-1}.n.(n-1)!}$$ giving your answer
$$ \dfrac {(2n-1)!}{2^{n-1}(n-1)!}$$