Prove that if n>=m then the # of m-cycles in Sn is given by [n(n-1)(n-2)...(n-m+1)]/m
The order of Sn is n!. We're counting the # of ways of forming an m-cycle, then divide by the # of a particular m-cycle.
The Attempt at a Solution
This problem doesn't seem too bad. Initially I was thinking about using the induction proof. But now that I have doubts the induction would even work after I looking at an example since we have to prove for any given n in symmetric group, the # of m-cycles in Sn = [n(n-1)(n-2)...(n-m+1)]/m for all m<=n.
For example if n=5, then
1) m=1, (# of m-cycles in Sn) = 5/1 = 5
2) m=2, (# of m-cycles in Sn) = 5*4/2 = 10
3) m=3, (# of m-cycles in Sn) = 5*4*3/3 = 20
4) m=4, (# of m-cycles in Sn) = 5*4*3*2/4 = 30
5) m=5, (# of m-cycles in Sn) = 5*4*3*2*1/5 = 24
Any helpful tips would be greatly appreciated!