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 Aug23 awarded Nice Question Dec17 accepted Relationship between ordered and binary trees Dec16 comment Relationship between ordered and binary trees @Isomorphism: I am looking for a different bijection. The post you linked is about a bijection for full binary trees, whereas I require a bijection for a particular number of left and right children. Dec15 asked Relationship between ordered and binary trees Dec13 accepted permutation with cycles Dec13 comment permutation with cycles Thank you very much! Dec13 comment Simon Newcomb's problem For 1. I somehow have to use that $A_{\emptyset}(t)=1$ and $\frac{A_S(t)}{(1-t)^{d+1}}=\sum\limits_{n=0}^{\infty}t^n\prod\limits_i\binom{n+‌​d_i-1}{d_i}$ Dec13 asked Simon Newcomb's problem Dec13 comment permutation with cycles Thank you very much for your answer. Unfortunately I cannot find the sketch of the proof you were talking about. Dec11 asked permutation with cycles Dec7 awarded Commentator Dec7 comment Generating function for binomial coefficients $\binom{2n+k}{n}$ with fixed $k$ I am also looking for a combinatorical proof of this identity. Nov27 awarded Scholar Nov27 accepted Combinatorial Interpretation Nov27 accepted Generating function for binomial coefficients $\binom{2n+k}{n}$ with fixed $k$ Nov26 comment Combinatorial Interpretation Thank you, Dennis. However, I do not have much experience with combinatorics so I am stuck with coming up with ideas for the other two. I know that 1.) it is choosing $s$ elements out of a $(n-m)r$-set and 2.) choosing $r$ elements from a $n-m$-set and then choosing $s$ elements from those. Still I dont see how this can be intersections of sets with bad properties. Nov26 comment Generating function for binomial coefficients $\binom{2n+k}{n}$ with fixed $k$ Thank you, but do you have a page number for me, please. Nov26 comment Generating function for binomial coefficients $\binom{2n+k}{n}$ with fixed $k$ $$\sum_{n=0}^k \binom{k}{n} \left( \frac{1}{2t} \right)^k (- \sqrt{1-4t})^{k-n-1} (2t)^{-k+n}$$ Nov25 comment Combinatorial Interpretation So $f(n)$ is the number of graphs on n vertices without isolated vertices because $2^{\binom{k}{2}}$ is the number of elements in the intersection of all sets of labeled graphs on $n$ vertices with an isolated vertex?! Nov24 comment Generating function for binomial coefficients $\binom{2n+k}{n}$ with fixed $k$ I expanded the left hand side to $\sum_{n=0}^k\binom{k}{n}\left(\frac{1}{2t}\right)^k\frac{-\sqrt{1-4t})^{k-n-1}}‌​{2t}^{k-n}}$ but that's it. I do not know how to proceed.