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I have this simple looking question: "In tennis tournament, there are $2n$ participants. So, there must be $n$ pairs for the first round. In how many ways can such a pairing be arranged??". Now, shouldn't the answer be $^{2n}C_2$? But the book says it is $1 \cdot 3 \cdot 5 \cdots (2n - 1)$. Where am I going wrong?

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    $\begingroup$ Check for the case $n=2$. There are 4 people, and the first person can be paired with any one the remaining in 3 ways. Corresponding to each such pairing, the remaining two are paired automatically. Extend the logic... $\endgroup$ – Macavity Apr 14 '13 at 13:10
  • $\begingroup$ Ok, the automatic pairing part - I was missing that! $\endgroup$ – Parth Thakkar Apr 14 '13 at 13:11
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Your attempt would be the total number of matches needed if everyone should meet everyone.

In this case, however, you want something else. The reason the answer is what it is is the following: Player number 1 has $2n-1$ potential opponents. Whoever his opponent is, the next unassigned player has $2n-3$ to choose from. And so on. All in all, you get $(2n-1)\cdot(2n-3)\cdots3\cdot 1$.

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