This is from C. C. Chen & K. K. MENG, Principles and Techniques in Combinatorics, p. 21.
Let $A$ be a set of $2n$ distinct objects. A pairing of $A$ is a partition of $A$ into 2-element subsets; i.e., a collection of pairwise disjoint 2-element subsets whose union is $A$. For instance, if $A$ is the set of 8 players $\{a, b, c, d, e, f, g, h\}$ as given above, then $$\begin{align*} &\{\{a,b\},\{c,f\},\{d,h\},\{e,g\}\}\\ \text{and}\quad&\{\{a,h\},\{b,g\},\{c,f\},\{d,e\}\} \end{align*}$$ are different pairings of $A$. We note that the order of the subsets and the order of the 2 elements in each subset are immaterial.
EXAMPLE 1.4.4. Let $A$ be a $2n$-element set where $n\geq1$. Find the number of different pairings of $A$.
Solution. We shall give 3 different methods for solving the problem.
Method 1. Pick an arbitrary element, say $x$, of $A$. The number of ways to select $x$’s partner, say $y$, is $2n-1$ (and $\{x, y\}$ forms a 2-element subset). Pick an arbitrary element, say $z$, from the $2n-2$ elements of $A\setminus\{x, y\}$. The number of ways to select $z$’s partner is $2n-3$. Continuing in this manner and applying (MP), the desired number of ways is given by $$(2n-1)(2n-3)\cdots5\cdot3\cdot1.$$
I cannot follow this proof. How to interpret the number $(2n-1)$ alone? Why isn't the number of ways in which $x$ (and $z$) can be chosen counted?
What matters here is how many pairs of elements we select.
One way to do this is to select two of the elements, then select two of the remaining elements, then continue selecting two of the remaining elements until all the elements are exhausted. At first glance, this method suggests there are $$\binom{2n}{2}\binom{2n - 2}{2}\binom{2n - 4}{2} \cdots \binom{6}{2}\binom{4}{2}\binom{2}{2}$$ ways to select the pairs. However, the order in which we select those pairs does not matter, so we have counted each selection of the same $n$ pairs $n!$ times. Hence, the number of ways of partitioning a set of $2n$ elements into $n$ pairs of elements is \begin{align*} &\frac{1}{n!}\binom{2n}{2}\binom{2n - 2}{2}\binom{2n - 4}{2} \cdots \binom{6}{2}\binom{4}{2}\binom{2}{2}\\ & \qquad = \frac{1}{n!} \cdot \frac{(2n)!}{2!(2n - 2)!} \cdot \frac{(2n - 2)!}{2!(2n - 4)!} \cdot \frac{(2n - 4)!}{2!(2n - 6)!} \cdots \frac{6!}{2!4!} \cdot \frac{4!}{2!2!} \cdot \frac{2!}{2!0!}\\ & \qquad = \frac{1}{n!}\frac{(2n)!}{2^n}\\ & \qquad = \frac{(2n)(2n - 1)(2n - 2)(2n - 3) \cdots (6)(5)(4)(3)(2)(1)}{n!2^n}\\ & \qquad = \frac{n(2n - 1)(n - 1)(2n - 3) \cdots (3)(5)(2)(3)(1)(1)}{n!}\\ & \qquad = \frac{(2n - 1)(2n - 3)(2n - 5) \cdots (5)(3)(1)n!}{n!}\\ & \qquad = (2n - 1)(2n - 3)(2n - 5) \cdots (5)(3)(1)\\ & \qquad = (2n - 1)!! \end{align*}
where $(2n - 1)!!$ is read $(2n - 1)$ double factorial.
Chen and Koh get this result more simply by observing that no matter which element you pick first, there are $2n - 1$ elements remaining in the set which can be paired with that element. Once you have selected that pair, no matter which of the $2n - 2$ elements you pick next, there are $2n - 3$ elements which can be paired with that element. Iterating that process until all the elements are exhausted, they find that the number of ways to partition a set with $2n$ elements into $n$ pairs of elements is $$(2n - 1)!! = (2n - 1)(2n - 3)(2n - 5) \cdots (5)(3)(1)$$
We do not care about the order in which the elements are picked, just which pairs we form. Thus, we care about which element is paired with $x$. If $x$ is the first element for which we pick a partner, there are $2n - 1$ ways to select the element which is paired with $x$.