Okay, for any given Sn, there are n elements in Sn {1, 2, ... m,..., n}, so we have n choices for the 1st element, then n-1 choices for the 2nd element, so on and so forth, and we have n-k+1 choices for mth element, etc. So there are total of n(n-1)(n-2)...(n-m+1)choices to form a m-cycle, to...