Number of different minimum spanning trees in the graph

Question

Let $G$ be a complete graph with $2n\ge6$ vertices. All edges in the graph have weights $2$, except the edges in the following cycles $$v_1,v_2,...,v_n,v_1, \qquad v_{n+1},v_{n+2},...,v_{2n},v_{n+1}$$ which all have weights equal to $1$. According to Cayley's formula, the total number of spanning trees in this graph is equal to $(2n)^{2n-2}$. I would like to know if there also is a simple way to find the total number of minimum spanning trees in the graph.

Answer 1

If this is incorrect, please tell me and I will delete.

In this new graph, the MST will have total edge weight $2n$, consisting of $(2n - 2)$ 1-edge and $1$ 2-edge. This is possible for example by taking $v_1 \to v_2 \to \ldots \to v_{2n}$. Clearly, it is impossible to do better, since the graph with only edges of weight $1$ is disconnected.

Now we can separate our MST into the $\{v_1, v_2, \ldots, v_n\}$ part, the $\{v_{n + 1}, v_{n + 2}, \ldots, v_{2n}\}$ part, and a single 2-edge connecting the two parts. Now we can count the number of MST in each of the two parts, which are both $n$, since such MST must only consist of edges of weight $1$. We then connect them with a single edge, which has $n^2$ choices. Therefore, in $G$ we have $n^4$ MST.