Question

​Let V be an arbitary vertex Kn. Find a formula for counting the number of spanning trees of G where v is a pendant vertex

Let v be an arbitrarty vertex of K pendant vertex Find a formula for counting the number of spanning trees of G where v is a

Answer 1

Theorem: (cayley, 1889) complete graph on n vertical has n^(n - 2) different spanning =recs theorem proof: Here we show that there is a 1-1 correspondence between the collection of all labelled trees on the vertex set (1, 2, ..n) and the collection of all 9n - 20-tuples of (1, 2..n). then the theorem will follow because the collection has cordinality n^(n - 2) First, we consider a labelles tree T on the vertex set < 1, 2,...n > , then the corresponding (n - 2).tuple -t will be formed as below. Among all rendent vertices in T. we consider the one, say v, whose label is the least and then we delete v from T. and note down this process for the tree T-v and get the second component at t, and so on until a single edges is left. No. of spanning trees = n^(n - 2) where n rightarrow no. of vertex.