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EDIT: This should probably be the most concise way of asking the question,.

Given $n,k\in\Bbb N$ such that $1\le$$k\le$$n$, how many permutations with order $k$ exist in $S_n$?

I'll elaborate further when requested.

xenophon
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  • How about two $k$ cycles? If $n=4, k=2$ does $(2,1,4,3)$ count? – Ross Millikan Nov 05 '14 at 01:20
  • I've been looking around for derangements and subfactorials, but it appears that is not how I should solve the problem at hand. I've altered the question so that it is now clearer what I seek. – xenophon Nov 07 '14 at 02:09
  • Now you have added more possibilities. If $k=4, n=8$ you can have a single four cycle, two four cycles, a four plus a two, or a four plus two twos. That is fine if that is what you mean. – Ross Millikan Nov 07 '14 at 02:19
  • I didn't understand what you meant with your comment, but perhaps I still lack clarity. When I talk of order, I'm referring to the number of times you have to apply said permutation in order to get back to starting (1,...,n) sequence. – xenophon Nov 07 '14 at 03:10
  • Yes, usually it is the minimum number of times you apply the permutation to get back. My first answer assumed you had one cycle of length $k$ and all the other elements were fixed, like $(2,3,4,1,5,6,7,8)$ for $n=8, k=4$. But there are also $(2,3,4,1,6,7,8,5), (2,3,4,1,6,5,8,7)$ and others. If you accept them, you need to consider the partitions of $n-k$ into divisors of $k$ and the formula will be more complicated. – Ross Millikan Nov 07 '14 at 04:36

1 Answers1

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If you require specifically one $k$ cycle, you can choose the elements in the cycle in $n \choose k$ ways. Then the first element of the $k$ cycle can go $k-1$ places, the one from where it went can go $k-2$ places, and so on to see there are $(k-1)!$ cycles available. The total number is then ${n \choose k} (k-1)!=\frac {n!}{k(n-k)!}$

Added: if you want any permutation of order $k$, there is no need to required $k \le n$. If $n=8$, you can have $(2,3,4,5,1,7,8,6)$ with order $15$. The maximum for a given $n$ is shown in A000793. For a given $k,n$ you need to find all partitions of $n$ into $k_1,k_2,\dots k_p$ with $k=\operatorname{LCM} (k_1,k_2,\dots k_p)$ For a given partition, the number of permutations is calculated as above: You select $k_1$ elements for the first cycle, $k_2$ for the second, etc., then choose the order of the cycle. This would give ${n \choose k_1}(k_1-1)!{n-k_1 \choose k_2}(k_2-1)!\dots {k_p \choose k_p}(k_p-1)!$ If some of the $k_i$s are equal, you have to divide by the factorial of the number of equal ones because you can reorder the selections.

Ross Millikan
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