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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?

3 Answers3

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A much neater way is to note that if the players are arbitrarily ordered, the first player has $2n-1$ possible matchups, the second unmatched player has $2n-3$ and so on. Thus there are $(2n-1)(2n-3)\cdots(3)(1)$ possible matchups, which is usually denoted as $(2n-1)!!$ using the double factorial.

Parcly Taxel
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That will be $\frac{2n!}{2^{n}n!}$

acat3
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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.

David K
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