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 Aug 23 awarded Nice Question Dec 17 accepted Relationship between ordered and binary trees Dec 16 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. Dec 15 asked Relationship between ordered and binary trees Dec 13 accepted permutation with cycles Dec 13 comment permutation with cycles Thank you very much! Dec 13 comment Simon Newcomb's problem For 1. I somehow have to use that $A_{\emptyset}(t)=1$ and $\frac{A_S(t)}{(1-t)^{d+1}}=\sum\limits_{n=0}^{\infty}t^n\prod\limits_i\binom{n+‌​d_i-1}{d_i}$ Dec 13 asked Simon Newcomb's problem Dec 13 comment permutation with cycles Thank you very much for your answer. Unfortunately I cannot find the sketch of the proof you were talking about. Dec 11 asked permutation with cycles Dec 7 awarded Commentator Dec 7 comment Generating function for binomial coefficients $\binom{2n+k}{n}$ with fixed $k$ I am also looking for a combinatorical proof of this identity. Nov 27 awarded Scholar Nov 27 accepted Combinatorial Interpretation Nov 27 accepted Generating function for binomial coefficients $\binom{2n+k}{n}$ with fixed $k$ Nov 26 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. Nov 26 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. Nov 26 comment Generating function for binomial coefficients $\binom{2n+k}{n}$ with fixed $k$ $$\sum_{n=0}^k \binom{k}{n} \left( \frac{1}{2t} \right)^k (- \sqrt{1-4t})^{k-n-1} (2t)^{-k+n}$$ Nov 25 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?! Nov 24 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.