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By ternary logic, the answer is "file not found."

​@@nisonaticdamn

Yes! The formula we derived here only works for m ≤ n, and for a variance computation you'll need the 2nd moment, i.e. m=2. So yes, the variance will be 1 for n at least 2. And in that one case when n=1, the variance is indeed 0.

Huh...I didn't even notice that. Just a coincidence--like how C(14, 6)=C(15, 5).

@@enumerable345 C(14,6) equals C(15,5) because 10 times 9 equals 15 times 6.

Isn't two enough? For sets A, B, C and bijections f: A -> B, and g: B -> C you have: - f^-1: B -> A - g^-1: C -> B - g•f: A -> C - f^-1•g^-1: C -> A

​@@masonskiekonto590 No, if you do it that way you just have composition of bijections and the overall bijection is, while a bijection, is "fixed" in the sense that it has to follow gf. E..g, if h = gf then h(a) = gf(a). In the trijection we allow for an additional permutation/shuffling. It's like adding a 4th set D and then what you do will work. That is, with a trijection we are allowing for any h(a) != gf(a) but h(A) = gf(A) (and h be a bijection). That is, all around the triangle we want free bijections. If one is determined by the other we don't have the complete degree's of freedom that one expects from a trijection. (e.g., where there is no preferred starting point) Basically if you set h = gf in your example then imagine that, h is fixed but we find all possible bijections f and g then there is an h for each one. Call it H = {h | h = gf and f,g are bijections}. This set H is what we can choose for the 3rd side. if you let r ~ s mean r and s are bijections(on the same set). Then we want h ~ gf rather than h = gf. (we loosen the requirements that the map of the individual elements must be equal and only require the global nature of a bijection)

Not sure how I feel about that nickname. 😂

I haven't fully worked it out, but I also figured it would be too much for this video. The counting argument would go something like... the trijection still cancels out most stuff. Then there are other mini-configurations we can put into trijection that also sum to 0 to cover more. Eliminate as many as you can this way, and the remaining configurations that don't balance all (this is a guess) have sign (-1)^(n/2). Count up the weights of anything that remains to get the RHS.

What a nice comment. Thank *you*, and keep counting!

Was this in the video? At any rate, pretty easy to just plug in these values into the formula and see it is true.

14 choose 6 = 14!/(8!*6!)=15!/(10!*5!) * (10*9)/(15*6)=(15 choose 5) * 1

(n k) = n!/(n-k)!k! (14 6) = 14!/(14-6)!6! = 14!/8!6! = 14!•15/8!5!•6•15 = 15!/8!5!9•10 = 15!/10!5! = 15!/(15-5)!5! = (15 5)

@@KrasBadan amazing. can you use this to construct an equivalence between the two sets

@@MichaelDarrow-tr1mn Now can you find an equivalence which will always add a card to the 5 if all of them are lower than 15, while "splitting" the 15 into two cards that add up to 15 if 15 is part of the 5 cards.

Well, it is kind of a jokey term, made up for this video. But-if you'll allow me to defend myself-I don't know that the Bourbaki group coined the word with the bi- prefix for that specific reason. An alternative, which generalizes better to tri-, is that there is a bijection between A and B if there's an injection from A to B and from B to A (Schröder-Bernstein), so a trijection would have a cycle of injections A to B, B to C, and C to A. Though this is motivated etymology on my part! :P

@@enumerable345 maybe we should worry about if that generalises to category theory properly too the equivalent of a bijection is an isomorphism, an injection a monomorphism, and a surjection an epimorphism, but it's not the case that a morphism that is both a monomoprhism and an epimorphism is automatically an isomorphism that makes me think that having a monomorphism from A to B and a monomorphism from B to A would not necessarily be the same thing as having a monomorphism from A to B that is also an epimorphism from A to B, but I don't know how to properly reason about this

man we all wanna reach samw height

Ooh, I don't think I've seen that identity before! I'll save coming up with the trijective proof for an upcoming plane flight. (Also--fellow Mudder?)

@@enumerable345 Yes I am a mudder, class of '13 :^) I shouldn't be too surprised you went to HMC, since you've clearly studied under Prof Benjamin.

almost forgot to mention, they're equal to 5!

The involutions match up some subset of the total set. Where they're not defined, we call those the exceptions.

Matching is so much easier than counting!

There are indeed tons of analogous identities/proofs with trinomial coefficients.

Yep, Manim for everything!

Remove logical combinatorics lol

Underrated 😂

So kind, thanks! I definitely have ideas for a few more videos.

That's a really cute problem! Thank you for sharing. Here is my solution, encoded with rot13. Gur qrfpevcgvba vf fhofrgf bs {1,...,a} bs nal fvmr jubfr nirentr vf na vagrtre, jurer gur fvta bs n frg vf qrgrezvarq ol gur cnevgl bs vgf fvmr. Gur vaibyhgvba, tvira n frg F jvgu nirentr n, vf whfg gb gbttyr gur cerfrapr bs n! Gur rkprcgvbany frgf ner gur frgf bs fvmr 0 naq fvmr 1. Gur fvtarq fhz bs rkprcgvbany frgf vf 1 - a, urapr gung vf gur jubyr fhz.

Pi is a very geometry-inspired number, so it's pretty easy to see geometrically. Imagine a unit circle circumscribing a regular hexagon. The circumference (2 pi) is larger than the perimeter of the hexagon (which is 6). This does require that a "straight line" is the minimum length curve between two points--which is a bit nontrivial. But just because it's geometric doesn't mean there's no combinatorial angle--I'll think about it!

It's "Boreal" by Asher Fulero. I believe I looped it and slowed it down by about 20% for my use. That song is available in the RUclips Audio Library, so you can use it in your own videos easily.

Yeah, that was supposed to happen--a sort of audiovisual joke. Thanks! I'm working on my next video, but there are a lot of new techniques I need to learn to animate it, so be sure to subscribe!