$(2n-1)(2n-3)\cdots 3\cdot1 = \frac{(2n)!}{2^nn!}$

Question

Here is a question from the book An Introduction to The Theory of Numbers by Ivan Niven.

Suppose that $\mathbb{L}$ contains $2n$ elements, and that $\mathbb{L}$ is partitioned into $n$ disjoint subsets each one containing exactly two elements of $\mathbb{L}$. Show that this can be done in precisely $$(2n-1)(2n-3)\cdots 3\cdot1 = \frac{(2n)!}{2^nn!}$$ ways.

My Question

I do not understand how did he moved to this expression $(2n-1)(2n-3)\cdots 3\cdot1$ since, up to my understanding, choosing $n$ subsets of $2n$ set with $2$ elements each is $\Pi_{k=0}^n {2n\choose 2}$. Furthermore, how $$(2n-1)(2n-3)\cdots 3\cdot1 = \frac{(2n)!}{2^n n!}??$$

Answer 1

First you select a pair out of $2n$ (there are $\binom{2n}{2}$ possibilities).

Then you select a pair out of the remaining $2n-2$ (there are $\binom{2n-2}{2}$ possibilities).

Et cetera.

That gives $\binom{2n}2\binom{2n-2}2\cdots\binom42\binom22=\frac{(2n)!}{2!2!\cdots2!}=\frac{(2n)!}{2^n}$ possibilities.

But like this every partition has been counted $n!$ times so we must divide by $n!$ to repair that.

This leads to $$\frac{(2n)!}{2^nn!}$$ partitions.

Answer 2

Note, that:

$$2^n n! = 2\cdot 4 \cdot 6 \cdot ... \cdot 2(n-1) \cdot 2n$$

Answer 3

Multiply by $2$, then by $4$, then by $6$, etc....:

$$1\cdot3\cdot5\cdot\ldots\cdot(2n-1)=\frac{1\cdot2\cdot3\cdot4\cdot\ldots\cdot(2n-1)\cdot(2n)}{2\cdot4\cdot6\cdot\ldots\cdot2n}=\frac{(2n)!}{2^n\cdot1\cdot2\cdot3\cdot\ldots\cdot n}=\ldots$$