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## Homework Statement

Prove that if n>=m then the # of m-cycles in Sn is given by [n(n-1)(n-2)...(n-m+1)]/m

## Homework Equations

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!

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