Is Pascal's triangle friendly??

Рет қаралды 12,958

3 жыл бұрын

We solve a nice problem involving binomial coefficients and Pascal's triangle from the "friendly Mathematics Competition".
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Пікірлер: 66

@pascalstriangle1843 3 жыл бұрын

Yes. Yes I am.

@Thaplayer1209 3 жыл бұрын

Pascal's Triangle your account was only made for this video?

@k_meleon 2 жыл бұрын

Hi friend :)

@jarikosonen4079 2 жыл бұрын

@@Thaplayer1209 its not possible? A hacker? Making fake accounts can not be supported in freedom or speech?

@andrewfucarino9613 3 жыл бұрын

3003 is the first number other than 1 to appear in two consecutive rows of Pascal’s triangle. It is in the 14th and 15th row. I knew that, and I knew it was because 1001,2002,3003 appeared in the 14th row, so I had the answer right when you stated the problem. Man, that has never come in handy before!

@Reliquancy 3 жыл бұрын

If you did that in person it would seem like a Ramanujan moment lol

@yakov9ify 3 жыл бұрын

Tiny mistake at 3:30, numerator should have (n-r+1) not (n-k+1).

@pascalstriangle1843 3 жыл бұрын

Yes

@jaehyunbaig1885 3 жыл бұрын

it's not a mistake, he chose k when he was talking about the n'th row

@ronaldjensen2948 3 жыл бұрын

@@jaehyunbaig1885 It is a mistake, he's used both k and r when he should have used one or the other. If the numerator is (n-k+1) then the denominator should have been k!. Likewise if the numerator is (n-r+1) the denominator is r!

@leif1075 3 жыл бұрын

@@jaehyunbaig1885 no like Ronald said below it has to be both k or both r..otherwise what the heck isbthe relation between k and r..if you stop at k minus 1 in the numerator, then the denominator has to be k factorial NOT r or anything else factorial..unless k equals r of course..

@jaehyunbaig1885 3 жыл бұрын

@@ronaldjensen2948 okay I understood ~~

@danielbranscombe6662 3 жыл бұрын

fun extension to try. Determine conditions for when this is possible for the general ratio a:b:c with a,b,c positive integers

@goodplacetostop2973 3 жыл бұрын

10:43

@twistedsector 3 жыл бұрын

You are speed

@sabokizi 3 жыл бұрын

thanks

@CM63_France 3 жыл бұрын

I like your multi-video with the "multi good places to stop", I can't remember , 6 or 9 videos fusionned?

@goodplacetostop2973 3 жыл бұрын

CM63 9 videos

@ThePharphis 3 жыл бұрын

I liked the answer and I bet that someone saw that part of row 14 and thought "hey, this would be an interesting problem"

@backyard282 3 жыл бұрын

does anyone else constantly admire his biceps?

@malawigw 3 жыл бұрын

Not me because I train bodybuilding :p

@cancer101thefundamentalsof4 3 жыл бұрын

This guy climbs at an elite level of rock climbing. In his video celebrating 50,000 subscribers, he mentioned he had climbed a few 5.14s which only a tiny handful of the climbers can do. So it makes sense that you admire his biceps since he can probably do one- fingered one-armed pullups.

@AshishKumar-lh6bg 3 жыл бұрын

Best maths teacher

@pokoknyaakuimut001 3 жыл бұрын

🥺😍

@ostdog9385 3 жыл бұрын

Damn right

@pupnoomann7866 3 жыл бұрын

such a refreshing little problem. easy, but very pleasing to see it unfold.

@jimskea224 3 жыл бұрын

That final result, in terms of the integers which the binomial coefficients reduce to, is quite surprising and rather cool.

@leif1075 3 жыл бұрын

Why would anyone think of defining Pascal triangle in terms of binomials or combinatorics..it's random and not smart..isn't there are a more intuitive logical way..like plugging in multiples of 3 seeing which ones can be made from some multipken of 2 since it has to be an even number and etc...

@oussemabouaneni992 3 жыл бұрын

@@leif1075 Dude, that's literally how it was defined throughout history before Pascal even existed...

@ffggddss 3 жыл бұрын

One fact worth knowing for this question, is that, in row n of Pascal's triangle, C(n,k), successive terms are obtainable from their immediate predecessors by multiplying by factors n/1, (n-1)/2, (n-2)/3, ... because C(n,1) = (n/1) C(n,0) , C(n,2) = ½(n-1) C(n,1) , C(n,3) = ⅓(n-2) C(n,2) , ... , C(n,k+1) = [(n-k)/(k+1)] C(n,k), etc. This is quickly verified by using: C(n,k) = n!/[k!(n-k)!] So in order to find consecutive terms that stand in the ratio 1:2:3, requires finding positive integers n and k for which (n-k+1)/k = 2, and (n-k)/(k+1) = 3/2 This is a 2E2U problem (2 equations in 2 unknowns). And the equations are even linear! n - k + 1 = 2k; n - 3k = -1 2(n - k) = 3(k+1); 2n - 5k = 3 ===> (n, k) = (14, 5) . . . Check: (n-k+1)/k = (14-5+1)/5 = 10/5 = 2 (n-k)/(k+1) = (14-5)/(5+1) = 9/6 = 3/2 √ [C(n,k-1), C(n,k), C(n,k+1)] = [C(14, 4), C(14, 5), C(14, 6)] = [14·13·12·11/4!, 14·13·12·11·10/5!, 14·13·12·11·10·9/6!] = [1001, 2002, 3003] It works!! And according to this method, it's the only solution. Fred

@factorization4845 3 жыл бұрын

Such a satisfying solution. 1001 is quite fascinating

@darreljones8645 3 жыл бұрын

I honestly thought the answer to the math question would be "no"!

@pascalstriangle1843 3 жыл бұрын

@@darreljones8645 You don't know all of my secrets 😏

@voteforno.6155 3 жыл бұрын

It's a product of 3 consecutive primes, too.

@craig4320 3 жыл бұрын

Does this generalize to the case where the top of n choose r has an element from any associative algebra? (complex numbers, etc.) I think so.

@mrl9418 3 жыл бұрын

Oh no do more Vertex algebras !

@lazaroychasoperez8882 3 жыл бұрын

I have a problem: Let a,b,c,d positive integers (0 isn’t include) such that a*b=c*d, show that: a=gcd(a,c)*gcd(a,d)/gcd(a,b,c,d) If you had time please I need help with this exercise. Thanks, I admire you a lot.

@caesar_cipher 3 жыл бұрын

gcd(a,c)*gcd(a,d) = gcd(gcd(a,c)*a, gcd(a,c)*d) .... distributive property = gcd(gcd(a², ac), gcd(ad, cd)) .... distributive property = gcd(a², ac, ad, cd) .... associative property = gcd(a², ac, ad, ab) .... using ab=cd =a*gcd(a,c,d,b) .... distributive property = a*gcd(a,b,c,d) .... commutative property QED summary of properties used: Commutative - gcd(a,b)=gcd(b,a) Associative - gcd(a,b,c)=gcd(gcd(a,b),c) Distributive - a*gcd(b,c)=gcd(ab,ac)

@lazaroychasoperez8882 3 жыл бұрын

Caesar Cipher 👍

@farrasalharits5966 3 жыл бұрын

If you want the short answer instead of watching 10min video: No, they're not friendly. I tried petting my pascal triangle once but it bites me really hard.

@pascalstriangle1843 3 жыл бұрын

Shut up. That's what you get for pretending to throw the Fibonacci sequence and then secretly keeping it. Tricking your Pascal's triangle will come back to bite you in the ass.

@iridium8562 3 жыл бұрын

Pascal's Triangle the fact that you are Pascal‘s triangle made this 100000 times better xD

@jkid1134 3 жыл бұрын

Another very similar approach would be to find the factor that transforms a choose b into a choose b+1, and then set this factor equal to the desired ratios and solve.

@leif1075 3 жыл бұрын

Why would anyone think of applying bnimomials orncombonstorics tonthebtriangle..it's so random

@MrRyanroberson1 3 жыл бұрын

binomial coefficients are famously described by pascal's triangle. you can see this play out if you start with the number 1 and keep multiplying by (x+1), such as with 11 (10+1): 1, 11, 121, 1331, 14641 (it breaks when any digit carries over, but the idea remains)

@torkjeldkjeldsen2527 3 жыл бұрын

is the parker square unfriendly?

@JB-ym4up 3 жыл бұрын

By symmetry of the choose function n=14,k=10 is the same as n=14,k=5.

@peterbrough2461 3 жыл бұрын

.. and the ratios would also reverse 3:2:1 .... but that's a different question😉

@jarikosonen4079 2 жыл бұрын

Great. I first thought its diophante with multiple solutions or no solutions. Also by symmetry (14 choose 10), (14 choose 9) and (14 choose 8) (in reverse order 3:2:1). I could not be sure if there is 2 results as the other one is 3:2:1 ratio, but it could be. Make another exploration with diagonal elements that would be easy. Can you find pi somehow in this Pascals triangle? Maybe no... What about question for ratio x:y:z, so that it appears at least twice in the pascal's triangle (left side only) and solve for x,y,z? No result, I guess... Best case maybe (45 choose 11):(45 choose 10):(45 choose 9) ~ (40 choose 10):(40 choose 9):(40 choose 8) ~ 1:3.6:3.2 = (x:y:z) Not possible, because this system of equations can not give more than one result, no matter what ratio is in concern...

@Eyalkamitchi1 3 жыл бұрын

So there's only one occurrence of this thing?! I would have though it would occur more in an infinite triangle. Does the ratio of consecutive entries diverge?

@MrRyanroberson1 3 жыл бұрын

well as it happens, it's two equations with two variables. it only happens once!

@pandas896 3 жыл бұрын

One question I wanna ask , is what is the use of Pascal's triangle

@Jim-be8sj 3 жыл бұрын

One use would be to expand (x+1)^8 quickly. The coefficients for the terms with powers x^8, x^7, x^6 ... are found on the triangle. Probabilities involving coin tosses live on the triangle. Many uses besides these as well.

@andywright8803 3 жыл бұрын

If you like Poker, Pascal's triangle can be used to calculate the probability of all the different possible hands

@VerSalieri 3 жыл бұрын

Never... will be a good place for you to stop making videos.

@franchello1105 3 жыл бұрын

This must be the only solution, right?

@AshishKumar-lh6bg 3 жыл бұрын

Sir please take problem from prermo India.

@ribhuhooja3137 3 жыл бұрын

He generally takes from higher levels than the P-RMO. Typically National level olympiads the analogue of which would be INMO in India. He even did an INMO problem once. PRMO would be a bit lower level

@rajvardhansinghsisodiya1095 3 жыл бұрын

@@ribhuhooja3137 are you an indian

@ribhuhooja3137 3 жыл бұрын

@@rajvardhansinghsisodiya1095 Yes, and (if it's not cancelled because of the pandemic) I plan to give the PRMO next year (because it is unlikely that it will happen this year, though if it does, I'll give it). Which is why all the olympiad problems that are solved are such a help (Not just that, this channel is a goldmine for anyone who wants to get into competitive maths).

@rajvardhansinghsisodiya1095 3 жыл бұрын

Ribhu hooja are you a member of this channel

@ribhuhooja3137 3 жыл бұрын

@@rajvardhansinghsisodiya1095 No. Just subscribed