Tessellate S.T.E.M.S. (2019) - Mathematics - Category A (School) - Set 5 - Subjective Problem 2

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):

  1. 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 \( j \in \{1, \cdots , n \} \).

  2. 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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The 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 \[ \frac{k(k+1)(\cdots)(k+n-1)}{n!} = \binom{k+n-1}{n}. \]

Patrick Corn - 1 week, 6 days ago

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