Given the following unlabeled graph (link), how many spanning trees does it have? The graph is basically four heptagons, each having exactly one common vertex with its top/bottom and left/right heptagon. For example, the upper-left heptagon has one vertex in common with the upper-right one, and another (different) vertex in common with the bottom-left one.
My Attempt:
I've tried simplifying the graph first. Instead of heptagons, I tried to use triangles. Drawing out the possible spanning trees, I believe there are a total of $14$ different (non-isomorphic) spanning trees in such a structure. (This is the simplified graph with triangles instead of heptagons.)
However, even with this simplified graph, I don't have much of an idea of how to count the number of spanning trees [short of actually drawing them all out, which I've done but a) I am not sure if I haven't missed any and b) drawing them is infeasible for the original heptagon graph]. I've thought about the fact that the spanning tree should have $8 - 1 = 7$ edges, and there are $12$ edges in total in the graph, so maybe $\binom{12}{7}$ ways to pick out 7 edges, then minus the 7-edged subgraphs that have cycles. But then again, finding the number of 7-edged subgraphs that have cycles is an entirely different problem.
Note that I'm completely focusing on the simplified, triangle version of the graph instead of the original heptagon version; because I believe that if I manage to solve this simplified case instead, the heptagon case wouldn't be too different. However, I'm completely stumped. Note that the matrix-tree theorem is not intended to be used in this problem.
