# In how many ways can $2n$ players be paired?

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?

• 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... – Macavity Apr 14 '13 at 13:10
• Ok, the automatic pairing part - I was missing that! – Parth Thakkar Apr 14 '13 at 13:11

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 $(2n-1)\cdot(2n-3)\cdots3\cdot 1$.