We call a bijection $\sigma: \{ 1, \cdots , n\} \longrightarrow \{1, \cdots , n \}$ a "permutation" of the set $\{ 1, \cdots , n\}$. Let $S_n$ denote the set of all the $n!$ permutaions of $\{ 1, \cdots , n\}$. In particular, for $k \geq 2$ a permutation of the form $\tau: \{i_1, \cdots, i_k \} \longrightarrow \{i_1, \cdots, i_k \}$, for a subset $\{i_1, \cdots , i_k \}$ of $\{1, \cdots, n \}$ is called a "cycle of length k" or simply a "k-cycle" if $\tau(i_j) = i_{j+1}$ for all $j \in \{ 1, \cdots, k-1 \}$ and $\tau(i_k) = 1$ (the definition of a "1-cycle" being the obvious identity). Then it is easy to see that every permutation may be written uniquely (upto order of cycles) as a product of disjoint cycles. Thus it makes sense to speak of the number $cyc(\sigma)$ of cycles of an arbitrary permutation $\sigma$ of $\{ 1, \cdots , n\}$. It may be quite interesting to note that for any (fixed) positive integer $k$ the expression

$\frac{1}{n!} \sum_{\sigma \in S_n } k^{cyc(\sigma)}$

is independent of any permutation in $S_n$ and may be expressed as a very simple closed form (in fact a single binomial coefficient) involving nothing but $n$ and $k$. Find this binomial coefficient and prove the relevant identity.

A Few Hints Towards a Possible Approach (Spoiler Alert, goes without saying):

Given nonnegative integers $k_1, \cdots , k_n$ such that $\sum_{j=1}^n jk_j = n$ can you count the number of elements of $S_n$ whose cycle decomposition have exactly $k_j$ cycles of length $j$ for all $j \in \{1, \cdots , n \}$.

You may need the infinite formal series:

$e^t = \sum_{i \geq 0} \frac{t^i}{i!} = 1 + t + \frac{t^2}{2} + \frac{t^3}{3!} + \cdots$

and

$ln(1-t) = \sum_{i \geq 1} \frac{t^i}{i} = t + \frac{t^2}{2} + \frac{t^3}{3} + \cdots$

This problem is a part of Tessellate S.T.E.M.S. (2019)

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## Comments

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TopNewestThe answer should just be $\binom{k+n-1}{n}.$

The sum can be written as $\frac1{n!} \sum\limits_{m=0}^n a_m k^m,$ where $a_m$ is the number of permutations in $S_n$ with $m$ cycles. Then $a_m$ is the unsigned Stirling number of the first kind $\begin{bmatrix} n \\ m \end{bmatrix},$ and there is an identity (also from the link): $x(x+1)(\cdots)(x+n-1) = \sum_{m=0}^n \begin{bmatrix} n \\ m \end{bmatrix} x^m,$ so plugging in $k$ for $x$ and dividing by $n!$ immediately gives that our sum is $\frac{k(k+1)(\cdots)(k+n-1)}{n!} = \binom{k+n-1}{n}.$

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