Видео

clean

I was looking for this solution for so long!

Using binary math, where "0" is blue, and "1" is red. 001 - 3 versions (001, 010, 100) 011 - 3 vers 000 - 1 vers 111 - 1 vers Assume all versions (ie: 001, 010, 100) amount to the same thing, so just handle the once case shown (ie: 001). So, for 011, like in the video, if you are the middle "1", and the right "1" raises their hand, and you see the left "0", then you know you are "1" (red). Also, for 001, if you are the middle "0", and the right "1" doesnt raise their hand, then you know you are "0" (blue). For 000, if no one raises their hand, you know you are "0" (blue). For 111, everyone raises their hand. According to the video, if no one solved it as above, after a few minutes, then you are "1" (red).

Johny Carson once played this game, although I am not sure they got the answer: ruclips.net/video/PKTk4coq05U/видео.html

Thank you. I lost sleep over this one last night.

But if you just put 1 in the first or the second box, the box that is left will be empty too? Could you elaborate on that?

helll yeah

@ 10:47 I would have used the substitution u = pi - x instead.

@ 5:28 You didn’t close the bracket

Helped me so much dude, thanks! Keep up the good work :)

Thank you, this is an excellent explanation

why couldnt the same logic be applied in saying "we just have to find 4 ways to place b"? Man this is STILL confusing. Im sorry but for me it's always been true; this the one branch of maths i hate the most. And i sooo want to love it, like you do. gods.

Thank youu

thank you that was helpful

thx a lot dude

This video was an absolute lifesaver for my exams, thank you so much!!

Glad someone actually took the time to explain the intuition behind and not just fucking algebra

would love to see more about stirling numbers, it'll be fun to see them in their generating function's context, and the thing you talked about linear algebra

That is sooo cooool

1★★★★★★

Thank you

It's actually 7 then 5 then back to 1 then it loops to infinity no matter how high the number gets. It's also how everything is made. Even you. 1+1 = mum and Dad 2+2 = grandma and Grandad and so on and so on it goes

Not only there is a recursive formula but also an explicit formula.

This was super helpful! I was trying to figure out this problem earlier (from another source), and even when looking at the solution I could follow it, but not figure out how to solve it myself systematically. Your method of looking at # outcomes vs # answers was super helpful, and let me understand the problem a lot better :)

I did it in a different way ruclips.net/video/XAlADhAgYAQ/видео.html

thank you so much this was so clear and really helped me understand it!!!

It was very helpful for me. Thank you.

Unfortunately, the quality is horrible, but the content is really interesting ...

great

@5:36, for instance, the coefficient for x^18 can be extracted from Wolfram Alpha as: SeriesCoefficient[3x/(1-x-2x^2),{x,0,18}]

Excellent presentation! You are the anti-math.SE in tone and friendliness!

Cool

4:33 shouldnt the no of items of arrangement be 5p3?

This is an excellent explanation, I could feel the lightbulbs going off in my head every ten seconds.

16 is next

Oh man This is way longer than expected

Fantastic explanation! Thank you so much for the effort. Really appreciate it

I like this guy's attitude.

Where you from

it was a good knowledge to get..upload videos like this pls.

Good 👍👍

Another cool, general way solving recursive relations is using matrices: *r_n := (x_{n-1}, x_n)^T ∈ ℝ^2 => r_n = A * r_{n-1}* with *A = (0; 1// 2; 1)* and the initial condition *r_1 = (0; 3)^T* . Via induction we get *r_n = A^{n-1} * r_1* though we need _Jordan's Matrix Transform_ to simplify the matrix exponential: *A = T * D * T^{-1}, | D = diag(2; -1), T = (1; -1// 2; 1)*

That was really good!

Can you do a video of THE EXTENDED BINOMIAL THEOREM with generating functions

Thank you, this was very helpful.

Great video; thanks!! So never seem to want to solve X, do we

i feel like one can also use this n choose k to practise magic

Thank you for this video dude! Appreciate your attention to detail in explaining the concept. Much help in solving my Combinatorics hw haha

the final problem is unexpected, but it helped me, many thanks to you!