Basic combinatorics question

Question

1. 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.

Answer 1

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

$$ \dfrac {(2n-1)!}{2^{n-1}(n-1)!}$$

Answer 2

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:

$$P(n) = \frac{1}{n!}\prod_{h=0}^{n-2}\frac{(2n-2h)!}{2!(2n-2h-2)!} = \frac{1}{n!2^{n-1}}\prod_{h=0}^{n-2}(2n-2h)(2n-2h-1) = $$ $$= \frac{1}{n!2^{n-1}}\frac{(2n)!}{2} = \frac{2n(2n-1)!}{n(n-1)!2^{n}} = \frac{(2n-1)!}{2^{n-1}(n-1)!}$$