334 Solutions to Exercises 2 -1 0 0 -1 2 -1 0 of spanning trees is the determinant of the (n - 1) x (n - 1) matrix 0 -1 2 0 0 0 -1 2 It is a straightforward exercise in mathematical induction to prove that this determinant is n. (b) There are n spanning trees not containing edge la (by part (a». Now suppose la is in some spanning tree. We must now delete exactly one of the edges 12,23, ... , (a -l)a and exactly one of the edges a( a + 1), . .. ,n1. There are a-I choices in the first list and n -(a -1) choices in the second. So the number of trees containing la is (a -1)(n -(a -1)) and the total number of trees is 1 n1J2 ... . 3 . . 0 •••••• a n + (a -1)(n -(a - 1)) = an -(a -1)2. (Alternatively, use Kirchhoff's Theorem!) 16. [BB] There are (~) possible edges from which we choose n -1. The number of graphs is, therefore, n No. of trees No. of graphs (~~1). The number of trees on n labeled vertices is nn-2. For n ::; 6, the table shows the numbers of trees as compared with the numbers of graphs.
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.