Number of spanning trees isomorphic to a specific tree

Question

I am trying to understand why this statement ($\frac{5!}{numof...}$) is true but I currently have no leads. explanations/hints appreciated.

Answer 1

Let $v_1,v_2,v_3,v_4,v_5$ be the vertices of $K_5$. Suppose that we want to embed the first of those three trees — call it $T$ — in $K_5$. There are $5!$ ways to assign the vertices of $T$ to the labelled vertices of $K_5$. For instance, we could make $v_3$ the central vertex (the one of degree $4$), $v_1$ the northwest leaf, $v_2$ the southeast leaf, $v_4$ the northeast leaf, and $v_5$ the southwest leaf. But as long as we make $v_3$ the central vertex, we get exactly the same subtree of $K_5$ no matter how we assign the four leaves to the vertices $v_1$, $v_2$, $v_4$, and $v_5$, and there are $4!$ ways to assign the four leaves to the vertices $v_1$, $v_2$, $v_4$, and $v_5$. The same thing happens no matter how we assign the central vertex, so each subtree of $K_5$ isomorphic to $T$ is counted $4!$ times in the original figure of $5!$, and as a result there are only $\frac{5!}{4!}$ copies of $T$ embedded in $K_5$. Finally, each of the permutations of the leaves amongst the vertices $v_1$, $v_2$, $v_4$, and $v_5$ corresponds to an automorphism of $T$: any permutation of the vertices of $T$ that leaves the central vertex fixed is an automorphism.