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

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In the symmetric group $S_n$, the number of distinct r-cycles is given by the formula ##\frac{n!}{r(n-r)!}##. This is derived from the fact that there are ##\binom{n}{r}## ways to choose r objects from n, leading to ##\frac{n!}{r!(n-r)!}##. To account for the cyclic nature of r-cycles, one object is fixed at the start, which eliminates duplicate cycles generated by cyclic permutations. The remaining r-1 objects can be permuted freely, resulting in a distinct r-cycle for each arrangement. Thus, the final count of distinct r-cycles is accurately represented by the formula.
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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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