Shouldnt it be (n!)/2 ? Equivalent to permutation of n distinct numbers except that we need to count each permutation once, since for any permutation, there would also be a reverse permutation that would result in an identical mst in the given scenario.
-- Dipit Grover B.Tech in Computer Science and Engineering - lllrd year IIT Roorkee, India -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.