I'm trying to find a combinatorial proof to show that for $1\leq r\leq\lfloor n/2\rfloor$ there are $$\frac{n}{n-2r}\binom{n-r-1}{r}$$ matchings of size $r$ in $C_n$.
I tried to argue as follows: given a matching of size $r$ in $C_n$, we can contract the edges from this matching, giving $C_{n-r}$ with $r$ distinguished vertices. Now $$\frac{n}{n-2r}\binom{n-r-1}{r}=\frac n r \binom{n-1-r}{r-1},$$ and given we could say that in $C_{n-r}$ the one distinguished vertex is labeled with $1$, after which there are $\binom{n-1-r}{r-1}$ ways to select the other distinguished vertices. After this, the selected vertices can be extended to edges from a matching, giving an $r$-matching in $C_n$. However, somehow I want to argue that there are $n/r$ ways to do this (on average, since $n/r$ need not be an integer?) but I don't really see how.
Any help or different approaches are much appreciated.
Take a fixed vertex from $C_{n-r}$ and assign it an index in $[n]$ - this vertex will not be in the matching.
From the remaining $n-r-1$ vertices, choose $r$ to be in the matching and expand them to edges to form a matching - number the vertices in the resulting cycle using the vertex we numbered at the start by going clockwise and incrementing the index.
For a given r-matching in $C_n$, in how many ways can we get it from the above process?
Notice that after choosing the index of the fixed vertex there is only one way of getting the matching - so we just need to choose an index at the start that correspond to a vertex in $C_n$ that does not participate in the matching. As there are $n-2r$ such, there are $n-2r$ ways to get the matching.
Thus, the total number of ways is $$\frac{n}{n-2r}\binom{n-r-1}{r}$$