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

We call a bijection σ:{1,,n}{1,,n} \sigma: \{ 1, \cdots , n\} \longrightarrow \{1, \cdots , n \} a "permutation" of the set {1,,n}\{ 1, \cdots , n\} . Let SnS_n denote the set of all the n!n! permutaions of {1,,n}\{ 1, \cdots , n\}. In particular, for k2 k \geq 2 a permutation of the form τ:{i1,,ik}{i1,,ik} \tau: \{i_1, \cdots, i_k \} \longrightarrow \{i_1, \cdots, i_k \} , for a subset {i1,,ik} \{i_1, \cdots , i_k \} of {1,,n} \{1, \cdots, n \} is called a "cycle of length k" or simply a "k-cycle" if τ(ij)=ij+1 \tau(i_j) = i_{j+1} for all j{1,,k1} j \in \{ 1, \cdots, k-1 \} and τ(ik)=1 \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(σ) cyc(\sigma) of cycles of an arbitrary permutation σ \sigma of {1,,n}\{ 1, \cdots , n\}. It may be quite interesting to note that for any (fixed) positive integer kk the expression

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

is independent of any permutation in Sn S_n and may be expressed as a very simple closed form (in fact a single binomial coefficient) involving nothing but nn and kk. 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 k1,,kn k_1, \cdots , k_n such that j=1njkj=n \sum_{j=1}^n jk_j = n can you count the number of elements of SnS_n whose cycle decomposition have exactly kjk_j cycles of length jj for all j{1,,n} j \in \{1, \cdots , n \} .

  2. You may need the infinite formal series:

et=i0tii!=1+t+t22+t33!+e^t = \sum_{i \geq 0} \frac{t^i}{i!} = 1 + t + \frac{t^2}{2} + \frac{t^3}{3!} + \cdots

and

ln(1t)=i1tii=t+t22+t33+ 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)

Note by Tessellate S.T.E.M.S. Mathematics
5 months, 2 weeks ago

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The answer should just be (k+n1n).\binom{k+n-1}{n}.

The sum can be written as 1n!m=0namkm,\frac1{n!} \sum\limits_{m=0}^n a_m k^m, where ama_m is the number of permutations in SnS_n with mm cycles. Then ama_m is the unsigned Stirling number of the first kind [nm],\begin{bmatrix} n \\ m \end{bmatrix}, and there is an identity (also from the link): x(x+1)()(x+n1)=m=0n[nm]xm, x(x+1)(\cdots)(x+n-1) = \sum_{m=0}^n \begin{bmatrix} n \\ m \end{bmatrix} x^m, so plugging in kk for xx and dividing by n!n! immediately gives that our sum is k(k+1)()(k+n1)n!=(k+n1n). \frac{k(k+1)(\cdots)(k+n-1)}{n!} = \binom{k+n-1}{n}.

Patrick Corn - 5 months, 2 weeks ago

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