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I am studying Problem 43, Chapter 10 from A Walk Through Combinatorics by Miklos Bona, which reads...

Let $A$ be the graph obtained from $K_{n}$ by deleting an edge. Find a formula for the number of spanning trees of $A$.

So how I approached this problem was by creating the Laplacian of A. I set the edge to be deleted as the edge between the first and second vertices in the graph. After an obscene amount of potentially dubious matrix operations, I received a result of...

$n^{n-3}*[2n^{3}-5n^{2}+3n \pm 1]$

Can anyone shed some light on this problem? I feel as I am approaching it the wrong way...

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There's no need to consider the Laplacian. We can obtain this by a simple symmetry argument.

Every edge of the complete graph is contained in a certain number of spanning trees. By symmetry, this number is the same for each edge, call it $k$. Let us now count the total number of edges in all spanning trees in two different ways.

First, we know there are $n^{n-2}$ spanning trees, each with $n-1$ edges. Therefore there are a total of $(n-1)n^{n-2}$ edges contained in the trees.

On the other hand, there are $\binom{n}{2} = \frac{n(n-1)}{2}$ edges in the complete graph, and each edge is contained in precisely $k$ trees. This means there are a total of $\binom{n}{2}k$ edges.

This gives us $$(n-1)n^{n-2} = \binom{n}{2}k$$ which upon simplification gives $k=2n^{n-3}$.

If we delete an edge, then we effectively remove the set of all spanning trees containing that edge. By assumption that number is $k$. Therefore there will remain $$n^{n-2} - k = n^{n-2} - 2n^{n-3} = n^{n-3}(n-2)$$ total spanning trees.

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  • $\begingroup$ Thank you very much for your very detailed and helpful response. I didn't previously consider counting all of the edges contained in the trees to be able to determine what to subtract from $n^{n-2}$ to receive my final answer. I guess I didn't need to use the Matrix-Tree theorem, which was the way I arrived at my original conclusion. $\endgroup$ – Scott Nov 20 '13 at 21:31
  • $\begingroup$ You're very welcome. I think an argument using the Matrix-Tree theorem will also ultimately pan out, but it'll be quite tedious. Either way, you know that your original answer cannot be correct since it's asymptotically larger than the total number of spanning trees before removing the edge. $\endgroup$ – EuYu Nov 20 '13 at 21:34
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Maybe an easier approach is to first count the number of spanning trees that containing that edge (which I think is equal to: $$\sum_{k=0}^{n-2} C(n-2,k) (k+1)^{k-1} (n-k-1)^{n-k-3}$$)

and subtract it from total $n^{n-2}$ spanning trees.

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  • $\begingroup$ Thank you for your response. Your method of getting to it is right (count the number containing the edge and remove it from the total number), but I don't think your summation is. The other answer gives the correct way (unless by some combinatorial magic it happens to work out the same). $\endgroup$ – Scott Nov 20 '13 at 21:33
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another approach:

Denote by $G_{i}$ the number of spanning trees on $K_n - e_i$ , where $e_i$ is the $i$-th edge in $K_n$ . Denote by $T_n$ the number of spanning trees on $K_n$ . We get the following equation wich I will explain bellow:

${\sum_{n=0}^{ n \choose 2} G_i} = ({n \choose 2} - (n-1))\cdot T_n$

Indeed any spanning tree of $K_n$ either contains the edge $e_i$ or it doesn't, so in the sum on the left hand side we actually count all spanning trees on $K_n$, but we are overcounting. When we evaluate $G_i$ we count all the spanning trees in $K_n$ evoiding $e_i$. Since every spanning tree has exactly $n-1$ edges, a spanning tree counted in the evaluation of $G_i$ (hence evoiding the edge $e_i$) is also evoiding ${n \choose 2} - (n-1)$ other edges of $K_n$, so we are counting it ${n \choose 2} - (n-1)$ times in the sum on the left hand side. By a symmetric argument we conculde that this is the factor by wich we are overcounting the spanning trees on $K_n$. Finally, again by symmetry note that the $G_i$'s all have the same value (call it $G$).

We get $ {n \choose 2}\cdot G = ({n \choose 2} - (n-1))\cdot T_n$.

Simplification yields $ G = \frac{(n-2)}{n} \cdot T_n$

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