Bipartite graph: Average number of cuts for a given partition Say I have a graph with an even number of nodes. We then randomly connect those nodes. The possible number of combinations boils down to $P=\frac{(2n)!}{2^n n!}$. We now make a partition of length $L$ on the graph, such that $L\in(1,n)$. This partition splits the graph into two subsets, and my question is: Given the randomly connected nodes, how many nodes will a partition cut? I attach a picture that I hope will helps understand the problem.
We define the average number of cuts as $Q(n,L)$. For n=2, we have a 4 nodes graph and there are $P=3$ ways to create edges. However, for the partition L=2, it will only cut the graph in two of the three possible cases. For n=3, there are 15 possible combinations, which I do not draw all of them.
I was able to almost find a general expression for it, $Q(n,L)=(P-x)L+xL$, where $P$ is defined before as the total number of combinations and $L$ is the partition. The problem is that I do not know how to express the factor $x$, nor do I know if this really generalizes.
It seems like a problem known for graph theory and may be related to cuts (https://en.wikipedia.org/wiki/Cut_(graph_theory)), but I am not sure.
Any help would be appreciated.
Edited: As asked in the comments, I attach a picture of all the n=3 possible combinations for L=3 and L=2 (L=1 is the trivial case). Notice how the geometry is a bit different, as they are no longer on a circle, but rather on two sets. 
 A: There are $L\times (2n-L)$ pairs of vertices such that the two vertices are on opposite sides of the cut. For each of these pairs of the vertices, $x$ and $y$, the probability that they are joined by an edge in the matching is $1/(2n-1)$; this is because $x$ is equally likely to be joined to each of the other $2n-1$ vertices.
Therefore, by linearity of expectation, the expected number of edges crossing the cut is
$$
L\times (2n-L)\times \frac1{2n-1}. 
$$
This is the expected number of cut edges. The quantity $Q(n,L)$ you are computing is something more like the total number of cut edges over all possible matchings. To go from expected to total, you multiply by the size of the probability space, which is the number of mathcings, equal to $$P = (2n)!/(2^nn!) = (2n-1)(2n-3)(2n-5)\cdots 1:=(2n-1)!!.$$ It follows that
$$
Q(n,L) = L\times (2n-L)\times \frac1{2n-1}\times (2n-1)!! = L\times (2n-L)\times (2n-3)!!
$$
This agrees with what you already found. For example,
$$
Q(3,3) = 3\times 3\times 3!! = 3\times 3\times 3=27\\
Q(3,2) = 2\times 4\times 3!! = 24\\
Q(3,1) = 1\times 5\times 3!! = 15
$$
For a further example,
$$
Q(4,3) = 3\times (8-3)\times 5!! = 3\times 5\times (5\times 3\times 1)=225
$$
