I am trying to find the number of matchings that are possible on graphs $G=(V,E)$ with $V = [n] = \{0, ..., n-1\}$
Let $a_n$ be the number of matchings on $[n]$ . Then I plan to calculate $a_n = \Sigma^{n}_{k = 0} b_k$ with $b_k$ being the number of matchings on $[n]$ with n nodes.
Then $b_k$ is $0$ for all not even k, since the number of nodes in a matching has to be even.
For even k: From the n nodes I want to select k nodes for the matching. So that there are $\frac{n!}{(n-k!)}$ ways to order in the selected k nodes. (n elements in the first position, n-1 in the second, ..., n-k+1 in the k-th position)
The first 2 elements then form an edge, the next 2 as well and so on. These form the matching. But it doesn't matter whether it is {a,b} or {b,a} , so I would divide by $2^{k/2}$
Also it doesn't matter whether the edge occurs at the first position or the last, so I would again divide by $(k/2)!$ so that since there are the options of ordering the edges.
This however doesn't really make a nice compact expression for using that in an generating function. The kind of binomial coefficient makes me wonder, whether to use the binomial theorem, but it doesn't quite match this.
I arrive to the same conclusion, using the fact that the number of perfect matchings in the complete graph $K_{2n}$ is $$\frac{(2n)!}{n!\ 2^n}$$
This can be shown by induction. If $a_n$ is the number of perfect matching in $K_{2n}$ then, you can pick a vertex $u$, match it with (2n-1) vertices, and then find a perfect matching in $K_{2n-2}$. Hence $a_n=(2n-1)a_{n-1}$, and $$a_n=(2n-1)(2n-3)\ldots 1 = \frac{(2n)!}{(2n)(2n-2)\ldots 2}= \frac{(2n)!}{2^n \ n!}$$
Then the $T_n$ total number of matching in $G$ is (with $\#K_{2k}$ the number of complete subgraphs on $2k$ vertices among $n$) \begin{align*} T_n &= \sum_{k=1}^{\lfloor n/2 \rfloor} \#K_{2k} \cdot \frac{(2k)!}{k! \ 2^k}\\ &= \sum_{k=1}^{\lfloor n/2 \rfloor} \binom{n}{2k} \cdot \frac{(2k)!}{k! \ 2^k}\\ &= \sum_{k=1}^{\lfloor n/2 \rfloor} \frac{n!}{2^k\ k! \ (n-2k)!}\\ \end{align*}
And I arrive at the conclusion then you. I don't think we can do much better though. I keep looking