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Cayleys formula states that for a complete graph on nvertices, the number of spanning trees is n^(n-2). For a complete bipartite graph we can use the formula p^q-1 q^p-1. for the number of spanning trees. A generalization of this for any graph is Kirchhoff's theorem or Kirchhoff's matrix tree theorem. This theorem looks at the Laplacian matrix of a graph. ( you may need to look up what that is with some examples). For graphs with a small number of edges and vertices, you can find all the spanning trees and this is often quicker. There are also algorithms such as depth-first and breadth-first for finding spanning trees.

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Q: How do you count spanning trees in a graph?
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Prove that a graph G is connected and only if it has a spanning tree?

Proving this is simple. First, you prove that G has a spanning tree, it is connected, which is pretty obvious - a spanning tree itself is already a connected graph on the vertex set V(G), thus G which contains it as a spanning sub graph is obviously also connected. Second, you prove that if G is connected, it has a spanning tree. If G is a tree itself, then it must "contain" a spanning tree. If G is connected and not a tree, then it must have at least one cycle. I don't know if you know this or not, but there is a theorem stating that an edge is a cut-edge if and only if it is on no cycle (a cut-edge is an edge such that if you take it out, the graph becomes disconnected). Thus, you can just keep taking out edges from cycles in G until all that is left are cut-gees. Since you did not take out any cut-edges, the graph is still connected; since all that is left are cut-edges, there are no cycles. A connected graph with no cycles is a tree. Thus, G contains a spanning tree. Therefore, a graph G is connected if and only if it has a spanning tree!


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