Timeline for Combinatorial proof of identity [closed]
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Dec 28, 2017 at 20:13 | comment | added | Alexander Burstein | @IgorRivin The first thing that immediately occurred to me is to take the usual stars-and-bars count with $N-n$ stars and $n+1$ bars and split it according to the position $j+1$ of the last bar. It is essentially equivalent to Zach Teitler’s answer. So I would call his answer “a usual proof”, if not “the usual proof”. In fact, it is so “usual” that I would propose to move this question to m.se. | |
| Dec 27, 2017 at 16:40 | vote | accept | Igor Rivin | ||
| Dec 27, 2017 at 11:12 | history | closed |
Pietro Majer Ben Barber Ben McKay Henry.L Wojowu |
Not suitable for this site | |
| Dec 27, 2017 at 11:11 | answer | added | Zach Teitler | timeline score: 5 | |
| Dec 26, 2017 at 18:14 | comment | added | Igor Rivin | @FanZheng You can call it whatever you want, but since obviously I had not seen it before asking the question (otherwise I would not have asked), I would not call it "the usual proof". | |
| Dec 26, 2017 at 16:56 | comment | added | Fan Zheng | @IgorRivin I'm just curious what do you think about the proofs given in the two comments above (which is what I call "the usual proof"). | |
| Dec 26, 2017 at 12:00 | review | Close votes | |||
| Dec 26, 2017 at 21:20 | |||||
| Dec 26, 2017 at 9:43 | history | edited | Martin Sleziak |
added (binomial-coefficients) tag
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| Dec 26, 2017 at 9:09 | answer | added | Aaron Meyerowitz | timeline score: 2 | |
| Dec 26, 2017 at 3:42 | comment | added | Igor Rivin | @FanZheng Please enlighten me on what "the usual proof" is? The one that first leapt into my mind was an induction proof (induction on $N-n,$ wherein the identity follows from Pascal's triangle). | |
| Dec 26, 2017 at 2:49 | comment | added | Fan Zheng | Probably the real gist of the question is the definition of "purely bijective proof". I'm wondering why the OP thinks the usual proof is not "purely bijective". | |
| Dec 25, 2017 at 22:32 | comment | added | Zach Teitler | You have $N+1$ reindeer and need to choose $n+1$ for your sleigh. They are ordered by redness of nose. You can choose one reindeer to lead the sleigh, call it the $(j+1)$th reindeer, then $n$ additional less-red-nosed reindeer to complete your team. | |
| Dec 25, 2017 at 22:28 | comment | added | Qiaochu Yuan | Split into cases based on the largest element of a subset of $[N+1]$ of size $[n+1]$. | |
| Dec 25, 2017 at 22:17 | history | asked | Igor Rivin | CC BY-SA 3.0 |