We call a bijection σ:{1,⋯,n}⟶{1,⋯,n} a "permutation" of the set {1,⋯,n}. Let Sn denote the set of all the n! permutaions of {1,⋯,n}. In particular, for k≥2 a permutation of the form τ:{i1,⋯,ik}⟶{i1,⋯,ik}, for a subset {i1,⋯,ik} of {1,⋯,n} is called a "cycle of length k" or simply a "k-cycle" if τ(ij)=ij+1 for all j∈{1,⋯,k−1} and τ(ik)=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) of cycles of an arbitrary permutation σ of {1,⋯,n}. It may be quite interesting to note that for any (fixed) positive integer k the expression
n!1σ∈Sn∑kcyc(σ)
is independent of any permutation in Sn 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 k1,⋯,kn such that ∑j=1njkj=n can you count the number of elements of Sn whose cycle decomposition have exactly kj cycles of length j for all j∈{1,⋯,n}.
You may need the infinite formal series:
et=i≥0∑i!ti=1+t+2t2+3!t3+⋯
and
ln(1−t)=i≥1∑iti=t+2t2+3t3+⋯
This problem is a part of Tessellate S.T.E.M.S. (2019)
#Combinatorics
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Comments
The answer should just be (nk+n−1).
The sum can be written as n!1m=0∑namkm, where am is the number of permutations in Sn with m cycles. Then am is the unsigned Stirling number of the first kind [nm], and there is an identity (also from the link): x(x+1)(⋯)(x+n−1)=m=0∑n[nm]xm, so plugging in k for x and dividing by n! immediately gives that our sum is n!k(k+1)(⋯)(k+n−1)=(nk+n−1).