Basic combinatorics question 
*

*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.

 A: 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
$$ \dfrac {(2n-1)!}{2^{n-1}(n-1)!}$$
A: Denote with $P(n)$ the number of pairing you are looking for. 
Suppose that two players are already coupled. Then the number of couplings is $P(n-1)$. There are $\binom{2n}{2}$ possibility to extract $2$ players from $2n$. Then:
$$P(n) = \frac{1}{n}\binom{2n}{2}P(n-1)$$
If we reiterate the previous formula, we have:
$$P(n) = \frac{1}{n}\binom{2n}{2}P(n-1) = \frac{1}{n(n-1)}\binom{2n}{2}\binom{2(n-1)}{2}P(n-2) = \ldots = $$
$$ = \frac{1}{n!}\prod_{h=0}^{n-2}\binom{2(n-h)}{2}P(1) = \frac{1}{n!}\prod_{h=0}^{n-2}\binom{2(n-h)}{2}.$$
where $P(1) = 1$, because when $n=1$, then there are only $2$ players.
Then:
$$P(n) = \frac{1}{n!}\prod_{h=0}^{n-2}\frac{(2n-2h)!}{2!(2n-2h-2)!} = \frac{1}{n!2^{n-1}}\prod_{h=0}^{n-2}(2n-2h)(2n-2h-1) = $$
$$= \frac{1}{n!2^{n-1}}\frac{(2n)!}{2} = \frac{2n(2n-1)!}{n(n-1)!2^{n}} = \frac{(2n-1)!}{2^{n-1}(n-1)!}$$
