<s>[INST] Chess tournament combinatorics There are $2n$ players in a chess tournament. The first round consists of pairing the players to
participate in $n$ matches with every player playing one match. In terms of $n,$ how many ways can
this pairing take place?
The first pair of people can be chosen from $\binom{2n}{2}.$
Since we chose the first pair, the next pair will be chosen from an amount of $\binom{2n-2}{2}.$ 
This pattern continues till $\binom{2}{2}$ which is for the $n$th pair.
Hence, the pairings will be $$\binom{2n}{2}\cdot\binom{2n-2}{2}\cdot...\cdot\binom{2}{2}.$$
Dividing by $n!$ gets the ways $2n$ people can be chosen to play $n$ games.
How do I simplify this?
 [/INST] After solving the problem (which you did correctly), you could  have simplified by observing that
$$ \binom{2n}{2} = \frac{2n(2n-1)}{2} = n(2n-1),$$
$$ \binom{2n-2}{2} = \frac{(2n-2)(2n-1)}{2} = (n-1)(2n-1),$$
and in general
$$ \binom{2k}{2} = \frac{2k(2k-1)}{2} = k(2k-1).$$
So the numerator you computed is
$$ n(2n-1)\cdot(n-1)(2n-3)\cdot(n-2)(2n-5)\cdots3(5)\cdot2(3)\cdot1(1).$$
In the denominator you have $n!$ canceling the terms
$n,$ $n-1,$ $n-2,\ldots,$ $3,2,1$ in the numerator, leaving
$$ (2n-1)\cdot(2n-3)\cdot(2n-5)\cdots(5)\cdot(3)\cdot(1),$$
which is the same formula you have seen derived in another way already.
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