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4 votes
1 answer
407 views

Inverse relationship between Stirling numbers of the first and second kind via generating functions

In combinatorics, a well-known result is that the matrix formed by the Stirling numbers of the second kind $\left(S(n,k)\right)_{n,k\geq 0}$ and the matrix of the signed Stirling numbers of the first ...
VerMoriarty's user avatar
4 votes
1 answer
379 views

Counting permutations with a fixed number of descents and an extra condition

I am computing the volumes of certain polytopes and it turns out that knowing a "closed formula" for the following number would help a lot. Determine the number of permutations $\sigma\in \...
Luis Ferroni's user avatar
  • 1,889
3 votes
2 answers
463 views

Ask for a reference or a proof of a combinatorial identity $\sum_{k=0}^n\binom{2n+1}{2k}\binom {k}{m} =2^{2(n-m)}\frac{2n+1}{2(n-m)+1}\binom{2n-m}{m}$

Could you please recommend a reference to or supply a proof of the following identity \eqref{combin-ID-Maclaurin}, or \eqref{first-equiv-form}, or \eqref{combin-ID-Mac-Equiv}, or \eqref{combin-ID-Mac-...
qifeng618's user avatar
  • 1,101
3 votes
2 answers
403 views

Closed form for product of Stirling numbers of the second kind

What does the following expression evaluate to: \begin{equation} \sum\limits_{k=1}^n \dbinom{n}{k} \cdot k! \begin{Bmatrix} n \\ k \end{Bmatrix} \cdot k! \begin{Bmatrix} n \\ k \end{Bmatrix} \end{...
MathsPower's user avatar