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# Spanning trees in K2,n watch

1. I got a question that states,

It is known that the number of spanning trees in the complete bipartite graph Km,n is given by the function f(m,n) where

f(m,n)= n^(m-1)*m^(n-1)

Prove that the formula gives the correct number of spanning trees for K2,n.

First of all I found that f(2,n)= n*2^(n-1)

Then i tried to come up with some sort of reasoning about the number of spanning trees, and would then show that it is equivalent to the above, but i'm going round in circles really. I may be going the wrong way about it completely, but not sure, any help would be much appreciated.

Thanks.
2. (Original post by King Ripper)
...
This may not be the best method, but:

Focus on the set with n elements.

Since each vertex in that set must be connected to a vertex in the set of two elements; how many ways can this be achieved?

You've now formed two connected components.

Consider the number of ways these two components can be joined by adding another edge, and the number of resulting spanning trees (not the same).

Once you've done that, I think you would also need to show that all spanning trees can arise via this method; although a consideration of the number of edges and the necessity when constructing the two connected components reduces the work to a minimum.

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Updated: March 26, 2011
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