(5) How many spanning trees does $K_{2,p}$ have (recall: this is the complete bipartite graph)
Hint: Let A and B be the two vertices on one side of the bipartite graph and 1,2,...,s
be the labels on the other side. For any spanning tree, argue that if $S_A$ is the set of vertices
connected to A and $S_B$ is the set of vertices connected to B, then $S_A \cap S_B$ can have at most
one element. Fix this element, say it is labelled i- then how many different spanning trees
are there? Argue this is now the same as the number of subsets of s-1.
To count the number of subsets a set of size p, here is one way to proceed: a subset could
be of size 0,1,2,..., p. How many of them are there of size k? It is
$\binom{p}{k} = \frac{p!}{k!(p-k)!} = \frac{p(p-1)...(p-k+1)}{k!}$
So the number in total is just $\sum_{k=0}^p \binom{p}{k}$. How do you evaluate this? Use the identity that
$\sum_{i=0}^p \binom{p}{i} = 2^p$ which you can verify from the binomial formula
$(x+y)^p = \sum_{i} \binom{p}{i} x^i y^{p-i}$
and then substitute x=y=1. The answer you find, namely that $2^p$ is the number of subsets
of a set of size p, has a simpler proof actually. Each of the p elements, it can be either in
the subset of not, so there are 2 choices for each element, and hence $2^p$ choices in total. So
you can avoid the binomial coefficients if you are not familiar with them (but they tend to
turn up a lot in graph theory!)
