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It is well-known that the number of acyclic orientations of $K_n$ is $n!$. Does anybody know of a combinatorial argument for this fact which uses the identity: $$n!=\sum_{k=1}^ns(n,k),$$ where the $... 4answers 1k views ### Stirling number identity via homology? This is a question about the well-known formula involving both types of Stirling numbers:$\sum_{k=1}^{\infty}(-1)^{k}S(n,k)c(k,m)=0$, where$S(n,k)$is the number of partitions of an$n$-element set ... 0answers 506 views ### A combinatorial bound involving Stirling numbers of the second type My previous question was solved in a very elegant way, hopefully this (seemingly more complicated) case is also easy for experts. I need the inequality$\Big(\prod^r_{i=1}p_i\Big)\sum^n_{j=0}(-1)^j\...
Let $S_{n,r}$ denote the Stirling number of the second kind. Define $A_{n,r}:=\frac{\binom{n+r-1}{n}(n+r)!}{S_{n+r,r}r!}$. I want to prove: \$A_{n,1}\ge A_{n,2}\ge..\ge A_{n,r}\ge \lim_{r\to\infty} ...