Proving ##\frac{n!}{r(n-r)!}## Distinct r-Cycles in $S_n$

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Homework Statement



In $S_n$, prove that there are ##\frac{n!}{r(n-r)!}## distinct r-cycles.

2. The attempt at a solution

I know there are n choose r ways to permute r out of n cycles thus ##\frac{n!}{r!(n-r)!}## but I don't know how they got ##\frac{n!}{r(n-r)!}##?
 
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Lee33 said:

Homework Statement



In $S_n$, prove that there are ##\frac{n!}{r(n-r)!}## distinct r-cycles.

2. The attempt at a solution

I know there are n choose r ways to permute r out of n cycles thus ##\frac{n!}{r!(n-r)!}##

Two r-cycles are the same if their cycle notations are cyclic permutations of each other. Having chosen our objects, we can avoid such multiple counting by fixing one object to appear the beginning of all the r-cycles. Each permutation of the remaining r-1 objects will then generate a distinct r-cycle.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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