Number of spanning trees on graph disjoint union.

Question

Consider a graph G on 200 vertices, created by adding the following edges $(v_1, v_{101}), (v_2, v_{102})$ to the disjoint union of two complete graphs $K_{100}$ with respective vertices $v_1,...v_{100}$ and $v_{101},...,v_{200}$. What is the number of spanning trees for graph G?

I know that $t(K_n) = n^{n-2}$ according to Cayley's formula.

But I'm not sure about the disjoint union, does that mean that after performing a disjoint union, I get a discontinuous graph with 2 components, where each component is a complete graph $K_{100}$, that is then made continuous by the edge addition?

And if that is the case, would the answer then be $t(G) = 2\cdot(t(K_{100})^2)$?

Answer 1

But I'm not sure about the disjoint union, does that mean that after performing a disjoint union, I get a discontinuous graph with 2 components, where each component is a complete graph $K_{100}$, that is then made continuous by the edge addition?

Yes. (Although the correct terms are disconnected and connected).

And if that is the case, would the answer then be $t(G) = 2\cdot(t(K_{100})^2)$?

No. There are two cases to consider: exactly one of the extra edges is present in the spanning tree, or both are present. Your expression corresponds to one of those cases.