how many rectangles are there on a chessboard

how many rectangles are there on a chessboard

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math, but FAST!

[ishikani]

And if anyone wonders if it can be done any other way, let the side length be n. The amount of rectangle in an n × n square is exactly 1³ + 2³ + 3³ + ... n³ = n²(n+1)²/4. :)

[anonymousyt]

As a chess veteran I am happy that chess is getting light.

[ZipplyZane]

I had to rewatch the beginning to get it. What you're arguing is that every rectangle on this board will just be a combination of four lines--two horizontal and two vertical. There are nine horizontal lines and nine vertical lines, so you can choose any 2 out of 9 with each. Two independent choices are multiplied by each other to get the full number of possibilities, so the result is 9 choose 2 times 9 choose 2.

[klausolekristiansen]

Twelve thousand twelve hundred and ninetysix

[e-learningtutor]

I didn't even get the idea of finding number of rectangles in chessboard..
Lmao
Anyways u r a wizard sir

[dhruvap]

x^x^x^x=√-1; what's the answer for x please say bprp ?:)

[artemis]

Why do u always hold a grenade while solving mathematics?

[binaryparrot]

I have never looked at this question using possibilities... genius!!!

[AceLordy]

Could you make a few videos on combinatorics ? The topic is so fascinating. Especially when it comes to solving problems related to geometry.

[LearnFirstEarnNext]

Please can you give, how many sudoku puzzles are existing?

[Dhruva_P_Gowda]

x^x^x^x=√-1 what's x ? Please reply or make a video please 🥺:)

[yamansinha]

I rather collect six other dragon balls and ask sheron the answer.

[SOBIESKI_freedom]

A suggestion for a math video:

"How long is a piece of string?"

[doublepositivezero]

It was intended to be an aid for psychological hostility, though it's got human element randomization, it's the same as saying "unknown 3d plain object unlogged".

[simonwillover]

Another way to write it is to sum up all of the rectangles with widths and heights within the bounds. The number of rectangles with a width m, and a height n, that can fit on the chess board is (9-m)*(9-n). You can sum the number of rectangles for every width and for every height like so:

sum from ( m=1 to 8 ) of ( sum from ( n=1 to 8) of ( (9-m)(9-n) ) )
LaTeX: \sum_{m=1}^{8}{ \sum_{n=1}^{8}{ \left( 9-m \right) \left( 9-n \right) } }

[Lol-kekg]

Teacher :- How many square is a chess board
Me :- Loading😂😂😂😂

[bigjazbo]

Another method is consider any two "nodes" where lines meet. There are 81 choose 2 of these, or 3240. Each pair of nodes defines a rectangle except if the nodes fall on the same horizontal or vertical line, in which case the nodes define a line segment. There are 18 × (9 choose 2) such segments, or 648. That leaves 3240-648=2592 node pairs that can define opposite corners of a rectangle. Divide this by 2 since every rectangle can be defined two ways by opposite corners and you get 1296.

[dirac]

I found this solution to be very satisfying

[MathSolvingChannel]

Don't forget to put +C in each square

[thesloth_rm]

Too many rectangles... That's the answer