Timeline for The hexagonal property of Pascal's triangle
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| May 16, 2023 at 10:14 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
http -> https (the question was bumped anyway)
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| May 16, 2023 at 6:07 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title
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| May 15, 2023 at 15:13 | history | edited | Zach Teitler | CC BY-SA 4.0 |
Replaced lost image with tex
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Feb 21, 2011 at 7:07 | comment | added | Daniel Litt | Ah, by a combinatorial proof you mean a bijective proof. I agree that this is interesting, now. | |
| Feb 21, 2011 at 6:54 | comment | added | Theo Johnson-Freyd | I would be interested in knowing if there is any combinatorial proof for GCD-type results. It should be a separate question --- you have asked two, really. Since divisibility of binomial coefficients can be a fairly sensitive thing (Sierpinski-triangle type behavior), I would be surprised. | |
| Feb 21, 2011 at 6:50 | comment | added | Theo Johnson-Freyd | @milcak: I don't know if you're alerted when edits are made to answers. I have modified my answer to dramatically improve it; I hope that you will agree that it is in every way a bijective proof, although it is still of the form $A\times C = B \times C$ for some $C$. | |
| Feb 21, 2011 at 6:10 | history | edited | milcak | CC BY-SA 2.5 |
added 154 characters in body
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| Feb 21, 2011 at 6:08 | comment | added | milcak | @Theo Yes yes and yes that is what my question should be. I will edit the question. | |
| Feb 21, 2011 at 6:05 | comment | added | Theo Johnson-Freyd | (continuation) My suspicion rests on seeing many deeply combinatorial proofs that rest on various forms of double counting. In general, I often cannot give an explicit bijection between sets $A$ and $B$. But I often can find a set $C$ and give explicit bijections between $A\times C$ and $B\times C$. That's what's done here, and in Mitch's proof. | |
| Feb 21, 2011 at 6:01 | comment | added | Theo Johnson-Freyd | I think you are asking for the following: you are asking for someone to exhibit a natural bijection between, on the one hand, the set of triples of (a choice of a subset of size $m-1$ from a set of size $n-1$, a choice of a subset of size $m+1$ from a set of size $n$, a choice of a subset of $m$ from a set of size $n+1$), and on the other hand some other similar set of triples. You could revise your question if that's what you're really after, but I suspect it may not exist. (continued) | |
| Feb 21, 2011 at 5:58 | comment | added | Theo Johnson-Freyd | @milcak: Of course, products are the same as "do this and separately do that", so you shouldn't worry to much about disassembly. I've given what is much the same proof below, but I don't know if you'll like it any better. | |
| Feb 21, 2011 at 5:36 | answer | added | Theo Johnson-Freyd | timeline score: 6 | |
| Feb 21, 2011 at 4:12 | comment | added | milcak | @Theo No, I do not get turned off by such things - on these sites people do not know each others background, and so I understand why he posted that. However, his proof dissasambles the triads of the hexagons. I understand how you can see it is combinatorial, I just want something that uses the entire triad to count something, somehow. | |
| Feb 21, 2011 at 4:10 | comment | added | Daniel Litt | @milcak: The proof has to use the definition of the binomial coefficient somehow... | |
| Feb 21, 2011 at 4:09 | comment | added | milcak | The proof starts by expressing the property in a formula. I find that none combinatorial right away. Furthermore, to establish the equality he uses the same arguement 3 times, namely, a variation on an in-or-out arguement. I do believe there is a more intrinsic justification of this property. I'm looking for something such as one triple in the hexagon counts ..., the other ..., without actually coming back to each binomial independently. | |
| Feb 21, 2011 at 4:06 | comment | added | Theo Johnson-Freyd | I'm not a user at math.se, so I'll leave my comments here. Mitch's proof is completely combinatorial, and I don't expect you will do better here. I think you were turned off by the opening sentence, because you thought Mitch was speaking down to you, but that's a mistaken interpretation: most mathematicians are very careful to state everything, even things that they assume the reader knows; this is all the more important on sites like MO and M.SE, because other people are invited to read the posts, and they may not know as much as you do but still learn from reading the answers. | |
| Feb 21, 2011 at 3:48 | comment | added | Daniel Litt | Mitch's proof on math.se is purely combinatorial--what about that proof does not suffice? | |
| Feb 21, 2011 at 3:46 | history | asked | milcak | CC BY-SA 2.5 |