Suppose a bipartite graph $g$ consisting of $2n(n-1),n\in\Bbb N,n>1$ vertices, is divided equally into two colors: red and blue, and is constructed as follows:
For example, $g$ for $n=3$:
If I choose a set of vertices exclusively from one color, I could find the set of vertices from the other color that is a neighbour to at least one vertex in my set. For example, if I choose the following solid red vertices from the graph $n=3$, the set of unique neighbours to these vertices are highlighted in solid blue:
In this case, there are five unique blue neighbours. How could I go about finding the number of unique sets of red vertices such that the number of unique blue neighbours is exactly $k,2≤k≤n(n-1)$?
I know that finding the number of unique blue neighbours depends on the location of each red vertex (corner, edge or interior) and the location of each vertex with respect to the others (directly horizontally- and vertically- adjacent vertices will share two blue neighbours, while diagonally-adjacent vertices will share one), but I'm having a hard time conceptualizing a general approach much more sophisticated than brute force.