$2 n$ players participate in a tennis tournament. Find the number of possible pairings for the first round.
For the first player, there are $2 n-1$ possible opponents. Then, it remains to form pairs with the $2 n-2$ remaining players. Therefore, we have the following recurrence relation

$$
P_{2 n}=(2 n-1) \times P_{2 n-2}
$$

and thus

$$
P_{2 n}=(2 n-1) \times(2 n-3) \times \cdots \times 3 \times 1
$$

which is the product of all odd numbers up to $2 n$. This can be written using factorials as: $P_{2 n}=\frac{(2 n)!}{n!2^{n}}$.