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Let $S_n$ denote the group of permutations of $\{1,2,3, . . . , n\}$ and let$ k$ be an integer between $1$ and $n$. I need to find the number of elements $x \in S_n$ such that the cycle containing $1$ in the cycle decomposition of $x$ has length $k$.

Thank you so much

Surojit
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    Choose the other $k-1$ elements in the cycle, order them, then do whatever you want to the $n-k$ elements. – Steve D May 08 '16 at 02:05
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    @SteveD: Is it $\frac{(n-1)!}{(k-1)(n-k)!} \times (n-k)!$? – Surojit May 08 '16 at 02:09
  • Almost... you're missing the ordering of the other $k-1$ elements in the 1-cycle. – Steve D May 08 '16 at 02:10
  • @SteveD: Should I divide the above number by k? – Surojit May 08 '16 at 02:13
  • In what situations would you divide by a number in counting? In what situations would you multiply a number in counting? If we have picked what the $k-1$ extra elements in the cycle containing $1$ are, but do not yet know the order of said elements, and are currently counting all of these once, are we over counting or undercounting? Should we then multiply or divide? By how much? Remember that the number of ways to arrange $k$ objects in a circle is $(k-1)!$. – JMoravitz May 08 '16 at 03:11

3 Answers3

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As SteveD hinted at in the comments above: break apart via multiplication principle.

  • Pick the $k-1$ additional elements which appear in the cycle containing $1$

$\binom{n-1}{k-1}$ choices

  • Fixing the position of $1$, pick how the $k-1$ other elements in its cycle are ordered in relation to $1$

$(k-1)!$ choices

  • Pick how the remaining $n-k$ elements are permuted without regard to cyclic structure

$(n-k)!$ choices

Convince yourself that every permutation such that $1$ is in a cycle of length $k$ is described and described uniquely via a sequence of answers to the above steps. Via multiplication principle then, the total number of such permutations is the product of the number of options available at each step.

$\binom{n-1}{k-1}(k-1)!(n-k)! = \frac{(n-1)!}{(k-1)!(n-k)!}(k-1)!(n-k)! = (n-1)!$

Check to make sure that this answer makes sense. What values of $k$ are possible? If we sum over all possible values of $k$, what is the total number of permutations? Does this agree with what we expect?

$n$ possible values for $k$, and for each we have the same amount of $(n-1)!$. Thus there are $n\cdot (n-1)! = n!$ total permutations, exactly what we expected would happen.

JMoravitz
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We can choose the cycle $(1,\ldots)$ of length $k\geq1$ in $(n-1)(n-2)\cdots\bigl(n-(k-1)\bigr)$ ways, selecting one number after the other in the desired order. We then can permute the remaining $n-k$ numbers among themselves in $(n-k)!$ ways. Therefore the total number of permutations with $1$ in a cycle of length $k$ is $(n-1)!$, independently of $k$. It would be nice to have a "direct combinatorial proof" of this fact.

Christian Blatter
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Here’s a combinatorial argument.

First observe that the number of permutations of $[n]$ in which $1$ is in a $k$-cycle is the same as the number of permutations in which $n$ is in a $k$-cycle. Let $\sigma\in S_n$, write $\sigma$ in canonical cycle notation, and erase the parentheses to get a string $p_1p_2\ldots p_n$; then $n$ is in a $k$-cycle if and only if $p_{n-k+1}=n$, so that $(np_{n-k+2}\ldots p_n)$ is a $k$-cycle of $\sigma$.

It is well known that the map that takes $\sigma\in S_n$ to a permutation $p_1p_2\ldots p_n$ by writing $\sigma$ in canonical cycle notation and erasing the parentheses is a bijection. Thus, the number of $\sigma\in S_n$ such that $n$ is in a $k$-cycle of $\sigma$ is the number of permutations of $[n]$ that have $n$ in position $n-k+1$, which is $(n-1)!$ independent of $k$.

Brian M. Scott
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