Math for fun, how many rectangles?
ฝัง
- เผยแพร่เมื่อ 13 เม.ย. 2018
- How many rectangles are there in a 8x8 chess board?
The geometry behind 1^3+2^3+...+n^3, see 6:00, • 1^3+2^3+3^3+...+n^3 an...
Proof by induction, • Prove by induction, Su... ,
a combinatoric problem, find the pattern, discrete math problem, number of rectangles on a checkerboard
Subscribe for more math for fun videos 👉 bit.ly/3o2fMNo
💪 Support this channel, / blackpenredpen
🛍 Shop math t-shirt & hoodies: bit.ly/bprpmerch. (10% off with the code "WELCOME10")
🛍 I use these markers: amzn.to/3skwj1E
![[TH] 2024 PMSL SEA GF D3 | Summer | มาร่วมเป็นพยาน กับวินาทีแชมป์!](http://i.ytimg.com/vi/JCYvEsFlwx8/mqdefault.jpg)
![[Full Episode] MasterChef Junior Thailand มาสเตอร์เชฟ จูเนียร์ ประเทศไทย Season 3 Episode 1](http://i.ytimg.com/vi/Qs5zdBqLn_8/mqdefault.jpg)
Online classes have actually kill my love to math and physics in a single semester, this channel has recover that in a single video
th-cam.com/video/kow8ijXyVMQ/w-d-xo.html
I get what you mean brother. Self study is the best 👍
plus the restrictions of method is killing me
Same
Same here
Math is actually very cool. We are forced to study for tests, LSAT, or vestibular here, and that creates lots of stress. That's why it's boring at school. But when we try to solve problems just for fun... it's actually pretty cool.
I agree with you sir. It’s especially fun when you get to find the patterns of how some math works and find out by yourself about something like demonstrated in this video.
@2D ANIMATOR I wanna point that out too. The system try to force those stuff in there own way for everyone where only, like 2 out of 10 people would like it but doesn’t care about the others.
Math school is too ez it's boring (Im Grade 7)
@@HuntingKingYT hmmm, you’re not there yet. It’s getting better and better.
@@HuntingKingYT yeah you havent been doing nothing yet
Easy formula :- (r)((r+1)/2) * (c)((c+1)/2)
Where (r) represents number of rows and (c) represents number of columns
Absolute legend
nice
General formula: you start with x * y available points, any of which can be chosen first. You can then choose (x - 1) points (any point except for the first one) in the horizontal direction, and then (y - 1) points in the vertical direction. To make a rectangle, you’re left with only one option, and since you have only one rectangle, regardless of which of the four points you choose first, you have
[x * y * (x - 1) * (y - 1) * 1] / 4
Plugging in x = y = 9, you get
9 * 9 * 8 * 8 / 4 = 1296
Thumbnail: How many retangles
Oon Han yes
Oon Han my kind of clickbait (and comment bait)
blackpenredpen hahahah sure
The answer is 0. There are no redtangles in the picture, only blacktangles.
but squares are rectangles.
I’m seeing AT LEAST 12 rectangles there
I mean... you're not wrong
Very insightful.
Good eye captain obvious
@Ignited TNT nah, it's over 9 000 !
I reckon there are at least 17 hamburgers there
0 rectangles, the lines aren't perfectly straight, and the angles aren't exactly 90 degrees.
S it is square of chess board 😂
But it was mad question 👏
😂
@Debarghya Ray Official squares are also rectangles
U should change your Channel name to Magical comment 😂
@Manik Ray A square is a rectangle with all its sides equal!
If you think about it in terms of increasing sizes and positions, it is really easy to model in python code:
n_rect = 0
for h in range(1, 9):
for w in range(1, 9):
for y in range(0, 8-h+1):
for x in range(0, 8-w+1):
n_rect += 1
## Result: n_rect = 1296
Translator killed me."we get 9 choose 2" wrote "we got nachos too..."
LOLLLLLLL
Dont worry it is too cold for ice cream, we dont got them too...
I don't really get what it means by "9 choose 2" would you be kind to elaborate?
@@joels.5318 In Australia, we get taught it nCr, so it's nCr(9,2). Though, we pronounce it as 9C2. It's a combination; you use it to calculate the total amount of ways you can make a selection, provided it doesn't matter what order you select them in.
@@calebmacpherson4017 cool in what year do they teach that ?
Did it another way,
1. Take a random point as a corner (81 possible)
2. Take another random point as the opposite corner (81-1-8-8 = 64 possible)
3. Count the permutation of taking 2 oppositing corner from a rectangle (2×2 = 4)
Then you get 81 × 64 ÷ 4 = 1296
Beautiful way man!
That's impressive
Very elegant!
There's a lot of smart peeps in the comment section I see....and here I thought that I was good, lol
Wat ?
To form a rectangle you need 2 vertical and 2 horizontal lines :
Number of vertical pairs of line * number of horizontal pairs
Binomial coefficient of 2 in 9 = 36
36*36 = 1296
Beautiful approach! I instead defined the row and column indexes that define a general rectangle and wrote the sum over the possible choices of them and quickly got to the same formula.
1≤i≤j≤N for rows and 1≤a≤b≤N for columns. Then I can write [Sum from i to N of (N-i+1)]^2, which is [N(N+1) - N(N+1)/2]^2, which is [N(N+1)/2]^2.
For mxn grid, the number of rectangles is:
mn(m+1)(n+1)/4
Heliocentric the sun in your profile picture is the same as the one in the flag of Argentina.
first point mn second point (m-1)(n-1). 4 variant build rectangle if you have 2 points. result mn(m-1)(n-1)/4 m,n - vertex
I got a very different answer, but it still works 😛:
n = side length
Sigma(x=1, n)( 0.5x(n^2+n) )
X and y are side lengths
Add up the first x multiples of integers 1 through y
@@georgeelsham But the answer is same. And don't forget that Combinations can also be obtained from binomial expansions (that's the reason why it has links with summation method)
This is becoming my favorite channel in TH-cam...
@ekam maan what are you laughing at
I like to watch redpen. So relaxing. Wish there were more teachers like he ...
it has become mine too
Sameeeee ❤️❤️😻
There are many more like him, like Eddie Woo, Numberphile, patrickJMT, minutephysics, Paradoz Alpha Community
For us Physics teachers , this channel is a gold mine 🤩
in Python:
n = 0
for i in range(9):
for j in range(9):
n += i*j
Here the two nines stand for the rows/columns+1
man i am from brazil, and we speak portuguese here, your english fits perfect in my understanding, i understand all your vídeos, you are doing a amazing work to humanity, god bless you a lot
I wrote a little program in C and got 1296:
#include
int main(void)
{
int i, j;
int result = 0;
for(i = 8; i >= 1;--i){
for(j = 8; j >= 1; --j){
result = result + (i * j);
}
}
printf("There are %d rectengles!
", result);
}
if you replace the numbers in the for loop you can calculate any combination =)
/* This code is short, faster and works even if no. of rows and columns are not equal. */
#include
main(){
int r,c,ans;
printf("Enter the no. of rows & columns");
scanf("%d%d",&r,&c);
ans=(r*c*(r+1)*(c+1))/4;
printf("/n %d",ans);
}
/* This code doesn't use a single loop whereas your code will have time complexity O(n²). */
Or if you use a formula: i * j * (i+1) * (j+1) / 4, you can answer the problem in constant time rather than quadratic time.
I solved it by using the square of a summation; [E(i=0,7) 8-i]^2. I figured this out after realizing that for each unit over one a rectangle had in one direction results in one less rectangle that can fit on that axis. This can be represented as a summation where the index represents the number of units over one in a rectangle's side, and the function is the length minus the index. Then to extrapolate to variations in both direction you just square the summation.
lol that’s exactly how i did it and was looking for this comment
Put the smiley dragon there, use the helicopter hat on it or the double blades on put it on its head on the middle then once it's crossed over the cross square then it can become a baby crocodile with a half z on its head to make it a lizard? Something like that anyway they're not fully decided this is why I blow hot and cold
Very simple demonstration of 1^3+2^3+...+n^3=(1+2+...+n)²
You develop (1+2+...+n)² : you obtain n terms like k² for k=1...n and terms i.j where i and j are different.
The important point is that you remark that you have only one time each term i.j where i and j are fixed. In particular, for each time that you have i.j, you also have one time j.i. Then you rewrite your sum like this:
(1+2+...+n)²=1²+2²+...n² + 2.sum (i.j, where 0
My solution is long if you write it down, but short if you just think of it (the only calculation I did "outside" my head was the final 36*36)
Let the upper-left corner of a rectangle be "the starting point" of that rectangle.
For each point on the grid we can count, how many rectangles "start" on it.
The most upper-left point is a starting point of 8*8 rectangles (8 possible heights and 8 possible widths).
Similarly, the point below it is a starting point for 8*7 rectangles (7 possible heights and 8 possible widths).
It's easy to conclude, that the points in the first collumn are starting points for 8*(8+7+6+5+4+3+2+1) rectangles.
The points in the second collumn are starting points for 7*(8+7+6+5+4+3+2+1) rectangles (the only thing, that changed is the possible width - from 8 to 7)
So the sum of all the collumns looks like:
8*(8+7+6+5+4+3+2+1)
+
7*(8+7+6+5+4+3+2+1)
...
+
1*(8+7+6+5+4+3+2+1)
=
(8+7+6+5+4+3+2+1)*(8+7+6+5+4+3+2+1)=36*36=1296
Konstanty I made the same :)
= (n*n+1)²
Did the same :)
Guess I complicated it a bit by using combination
@@rasmussuonio3014 your fire spirits are explosive.(in your profile picture)
What you did is exactly what the combinatorics does. In fact using the combinatorics is the same logic described in that formula using factorials - which are the number of ways of choosing something from something.
Wouldn't the whiteboard make 1297 rectangles?
We can't see all 4 borders of the whiteboard so it could be a circle :P
So, how many rectangles are there once he is finished? (;
Daniel Gonzalez then wouldn't the screen make it?
yeah, and the screen you're using to watch the video would be the 1298th rectangle (so long as it's not the iPhone X)
you forgot to count the video rectangle frames in the recommendation section.
I really love your videos & explanations. Please do a series on Combinatorics. It's the scariest part of Mathematics.
This is beautiful. So many patterns in even the most simple math problems. I love math.
Yo dawg, I heard you like combinatorics, so we put a rectangle in yo rectangle so you can calculate while you calculate.
aint that kinda old?
QUALITY COLLECTOR think that’s part of the joke
@@user-gt7jk3lt4t it's an older meme but it checks out
LMFAO!!!!!!!!!!!!!!!!!!!!!! HILARIOUS!
Good video! It's cool to see that n(n+1)/2 here as well. I discovered it for myself about 9-10 years ago, but in a different context. One day, I was just doodling in the car, drawing uneven convex polygons, connecting all the vertices to every other one. I happened to start counting one of them, and soon enough it occurred to me that there could be a pattern, a connection between the number of vertices and the number of edges. I was determined to work it out. I drew polygons all the way up to the decagon (not bothering with uniform side length, knowing that it wouldn't affect what I was studying) and meticulously counted all the edges. I set out a table, and literally brute forced the pattern. I used the difference method and crunched it out. I then used my result to predict the number of edges for an 11-gon, and sure enough, upon actual inspection it was a match.
The best thing was, I wasn't a maths person at all up until that point. I knew the basics, as everybody does, and a rough grasp of algebra, but that was the extent of my abilities, and I think they were only that good because of formulae we had to learn for science, which I liked and was good at. I didn't know that what I was doing was an established thing. I didn't know what technique I was using; it just seemed logical to me that if I was looking for a pattern linking the number of sides and the number of edges, then I should look at the differences between consecutive totals. And repeat, until a constant term is reached. I did that, got the answer, and it was satisfying beyond words. I realized that I'd done maths, and not only was it fun, but it was kind of easy! This from a guy who was always in the struggling section of the maths class.
That was the start of my true mathematical education. It's far from over, of course, but I've come a long way from knowing just basic facts about geometry (triangle angles sum to 180, Pythagorean theorem, circle circumference, area, etc) and algebra. I'm confident with calculus material now, and my thinking in geometry and stuff related to it has radically evolved. Even my powers of visualization have improved an immense amount (I had just about zero when I started; now, I can hold and manipulate fairly complex statements.) I love maths now, and I always learn more of it whenever I can, and whenever I can't, I do what I mentioned - visualize. I've even thought about ways to generate mathematical objects via physical means, mechanical or otherwise, as cool devices/knickknacks that would be educational from a different perspective.
In any event, I've gone from moderate dislike/ambivalence towards maths to wanting it to be the foundation of my career, more than anything. It was a welcome surprise to have a vivid reminder of the catalyst that made that happen :)
This is such a motivational story!
@@mensch5502 thank you! It was all as I said :) but I guess I forgot the punchline - anybody can get good at maths!
@@marcushendriksen8415 that's what maths does : developing curiosity in those who learn it
KI head cover use the Pebble thick paint brush?
I would just divide cases
For 1x1 there are 8*8 = 64 rectangles
For 1x2, consider vertical ones : if you think of their upper component, there are 8*7 = 56 possibilities (whole grind w/o the last row). Same for horizontal ones, so there are 8*7*2 = 112 1x2 rectangles
in total, for 1xn rectangles there are 8*8 + 8*7*2 + 8*6*2 + ... + 8*1*2 rectangles possible which sums to 512 rectangles. 512 + 64 = 576
For 2x2, consider the upper left corner, it can be in a 7*7 grid of possibilities. Same for 3x3 rectangles which top left corner is in a 6x6 grid, etc.
In total for nxn, 7*7 + 6*6 + ... + 1*1 (we already counted the 8*8 1x1 rectangles) = 140. 576+140=716.
Then, 2x3 is 7*6*2 (horizontal and vertical), 3x4 is 6*5*2, etc. 7*6*2 + 6*5*2 + ... 2*1*2 (we already counted 1x2) = 224. 716+224 = 940. Here is my answer :D hyped to see if i'm right
edit : fuck
Really nicely done. Converting a varying rectangle problem into a combinatorics problem. Didn't see that at all. You explained it nicely.
The more simpler can be in horizontal choose any two lines and in vertical choose any two line
9c2 x 9c2 and the result is 1296
Good!
Good solution! I chose two points on the grid, and removed those that are on the same vertical line, then those that are on the same horizontal line (I did it for an n*m rectangle) and halved the result (since there are exactly two versions of every rectangle: the one with the top left and bottom right point, and the one with the top right and bottom left).
A smell a jee advance aspirant
I learnt this in vibrant academy MC sir! 😭
nice one!
My god i actually got it. Figured that there were (8+7+6+5+4+3+2+1) rectangles in each column. So i just squared it.
Whatnis the logic of doing that?
@@rahimeozsoy4244 he saw other comments.
Thats actually very neat way of thinking
every column has 36 & every row has 36 rectangles. So 36^2=1296 is the answer.
Damn that’s really clean
You make math more fun. I always want to learn more and each of your videos does that job perfectly. Now I have a request and I would love if you could accept it whenever you read this.
In some cases, I need to know what method I need to use to solve a mathematical situation that doesn't specifically ask a question. For instance, there is a picture of an object and it has included some text that hints me something mathematical about the object. Now my job here is to understand the whole picture, come up mathematical problems that are related to the picture, then solve them with formulas/methods. I am the one who creates the questions and show how much math I can do that is related to whatever the object is. To do that, I need to understand the point behind the picture and what part of math I can use for that specific picture.
Basically, I want to learn how to use math for anything I do or see in my everyday life. I see many people who only want to learn math because it's obligated to them. It's a subject that they must work with in school. I used to be one of them, but I have found my passion in math. I see math as a skill and not as a subject. I want to master math just for fun. I want to use in my everyday life, create math problems out of anything in the world and solve them no matter what situation I find myself in. It's hard to explain, and I'm not sure if there are many people out there who want something similar to me. That's why I'm reaching out to you.
I'm just an individual asking for some help from a genius teacher like you. If I'm correct, you always find ways/math-routes that works perfectly with the problem given. If you could make a video where you share your help for what I listed in this comment, I would greatly appreciate that.
I did it like this: I discovered that rectangles can be described as pairs of intersection points on the grid. Exactly 4 combinations map to each rectangle. For every point from the 81 we can choose a point that is not in the same column or row, so we don't just get a line segment. There are 17 (2*8 + 1) of them, that is the 1st point + 8 other points on the same row and column. Now we take 81*(81-17) and divide it by 4 and we get 1296.
pls do more vids on combinatorics..
I will!
It’s the number of ways of choosing a contiguous length from each side. For each side, you choose a length between 1 and 8, and then you can shift it up depending on the length you chose. So it would be the sum of (9-i) for i between 1 and 8, and then square the whole thing. Equivalently, it’s 8*9 minus the sum of i from 1 to 8, which equals 8*9 - 4*9 = 36, then squaring that we get 1296
EDIT: We could have also skipped a step by noticing that the sum of 9-i for i from 1 to 8 is the same as the sum of k for k from 1 to 8, just in reverse.
I did something similar
Owl square?
I like the various methods that you use. When you asked to solve it, I almost had it by the time you gave the answer. My logic was to work from the top down and develop a formula quickly. I saw that the 1x1 series was just 8x8 and the 2x1 series was just 7x7, but I forgot to calculate the Nx2 series and ended up with 1^2+2^2+3^2…+8^2 as my answer. And this is pretty easy to recognize because it effectively reduces the square size by 1 dimension. What I need to do was also recognize that for 3x2 and for any subset of squares that is not Nx1, you subtract the number of squares in the row by the n value of each side, add 1, and multiply by the other side. For example, an 8x8 square has (8-3+1)*(8-2+1)=6*7 sets of 3x2 squares in it. Then, you could either do some math to rearrange the terms and brute force the sum of the cubes or you could extrapolate the ((n(n+1)/2)^2) formula from the information.
Just commenting this because it’s also very interesting to do it from the top down and it’s still just as possible.
Thank you for this amazing channel, I really started to love math after your videos
I paused at 0:46 and i think i worked it out. You can view them as groups of rectangles that are all the same dimensions but in different spots. How many rectangles are in a group? Well there are as many spots horizontally as 9 - (width in squares). You then do the same thing vertically for each spot it can be horizontally. Or in other words you multiply (9 - [width]) by (9 - [height]). To get how many rectangles are in one group. And then you need to sum up all the groups. But summing every possible pair of valid height and width is equivalent to multiplying the sum of all possible heights with the sum of all possible widths. So (8+7+6+5+4+3+2+1)* (8+7+6+5+4+3+2+1) or 36*36 or 1296
We may state that each rectangle is defined by two opposite points, which are vertexes of the squares we begin with and the opposite vertexes of the rectangle. There are 9*9=81 points. If we choose ine point, we can't choose a point from the same line, so the second point is one of 8*8=64 remaining. Of course the first and the second points are interchangable and there are two pairs of defining points for each rectangle, so there are (81*64)/4=1296 rectangles.
Α combinatorics problem. For each rectangle we need 2 horizontal lines, chosen from a set of 9 and 2 vertical chose from a set of 9 as well. In general (N,M) = N!/(M!*(N-M)!) - where (N,M) means the ways we can chose M objects from a set of N. This gives [9!/(7!*2!)]^2=36^2. But it is real unbelievable how many ways are for solving this problem. Thank you blackpenredpen.
I'm glad that he touched on the combinations method, as that's how I first thought to solve this problem.
I feel so smart I knew exactly what was going on. Yay.
Your channel and mind your decisions channel are very helpful with me learning maths.
I did the math fast and got the right answer, but this fast solution was faster and an unusual perspective for me.
For the 1^3+2^3+3^3...+8^3 you can use the formula (1+2+3...+n)^2 = 1^3+2^3+3^3...+n^3.
I didn't really expect this would become my favorite channel in youtube😍
The fast solution blew my mind lol
XDBanküberfall laalDX 😊
XDBanküberfall laalDX it's exactly what I came up with when I tried this problem because I am lazy and I didn't really want to count rectangles
Hey, BPRP your channel is about to hit 100k subscribers!
sa fssfs I know!!!! I can't believe it!!!
@@blackpenredpen 690k now
Please believe 😁
Just learned about sums in algebra 2, so I was able to figure out the amount of squares: sum(n=1 to 8, n²)
From there I was pretty easily able to find the amount of rectangles: sum(l=1 to 8, sum(w=1 to 8, lw))
Very satisfying, and cool to use something I just learned :D
Just what I thought 🤝
i am very happy, i have never been able to solve your other puzzles but i tried this one. and surprisingly i found out the same formula i.e. (n(n+1)/2)^2. thank you for this puzzle, was really fun solving it!
but how to calculate how many squares with combinatorics? you'll choose (9 2) for one side but how many for the other one?
In total it would just be (9 2), because once you've chosen one side, the other side must be the same length, so there's only one choice.
Please do some more videos on permutation and combination 😹 that is legit the best part in mathematics
I'll assume that ur a jee mains/advance aspirant
I found another solution which is kind of nice. It’s a generalisation of the brute force method:
For a rectangle of size x by y, we want to count how many of them can be formed in the 8x8 grid. This is pretty easy - for example, a 3x2 rectangle can be placed in 6 horizontal positions and 7 vertical positions, for a total of 42. With some thinking a pattern emerges - there are (9-x)(9-y) ways to place a x by y rectangle in the 8x8 grid. The last thing to do is just double sum this expression for x and y between 1 and 8. Each sum just generates a constant of 36, which when squared is of course the desired answer - 1296.
I love how he is like "gtfo of here" in thumbnail
Funny that you use triangles numbers for calculating how many rectangles you have. Triangles numbers are : 1; 3; 6; 10; 15; 21 and so on
I read this thinking it said “how many triangles?”
I clicked hoping he would show me how he gets triangles from squares.
Sadly that is not the case.
cut it in half
Holy shit, same lol
We need to rest I guess
Hyperbolic spacial planes
@@johnnyknight6447 you mean spherical?
In spherical space you can make a triangle with 90 degree angles.
Simply choose any two parallel lines in each direction by 9c2*9C2
"Computing students call it divide and conquer "
*Shivers in dynamic programming PTSD*
What?! The partial sum of the cubes is equal to the partial sum of the whole numbers squared?! My mind was just blown
Yes : )
This is pretty basic stuff actually. We know this from 9th grade.
There's a very cool image about this on this page:
en.wikipedia.org/wiki/Squared_triangular_number
@@md.mohaiminulislam9644 shut up
@@LogicRick well he is right, if your preparing for ntse than your teacher will just tell tricks and logics for solving mat ques and it was one of them and pretty basic
I did a similar yet different approach than the fast solution.
Looking at the grid as a matrix, each cell stands for a unique type of rectangle that can be found on the grid with subindexes being their base and height: a(1,1) smallest square, a(1,8) column, a(8,1) row, a(8,8) the outer square, etc.
If the value of each entry is the number of rectangles of that type that can be found on the grid, the sum of all entries of the matrix is the number of rectangles on the grid.
The general way of computing the number of rectangles a(i,j) on the grid is (# of steps it could be moved horizontally +1) · (# of steps it could be moved vertically +1). That is (8-i+1)(8-j+1) = (9-i)(9-j).
So, the answer is Σ(Σ(9-i)(9-j)(i=1 to 8))(j=1 to 8).
And because 9-x from 1 to 8 is the same as x from 8 to 1, the answer simplifies as ΣΣij (i=1 to 8)(j=1 to 8).
ΣΣij (i=1 to 8)(j=1 to 8) = Σj·(Σi (i=1 to 8))(j=1 to 8) = Σj·36 (j=1 to8) = 36Σj (j=1 to 8) = 36·36 = 1296
Wow, very nice!!
Why are there so many smart people in this comment section? Makes me feel a lil bad about myself lol
This is first video of this channel that TH-cam recommended me.. I saw it for a few minutes & subscribed immediately
Now I commenting this after watching the full video.
Vertical sides: C(9,2)
Horizontal sides: C(9,2)
36^2=1296
Now I understand why we learn patterns and differences of numbers in lower classes...😎😎
9c2 *9c2 is the first thing that came to my mind when i first saw the thumbnail
Does this hold true for every rectangular shape?
I think it does! Thanks for your comment
That puzzle was pretty fun!
Was a good challenge to do without paper, but still really straightforward and simple.
Figuring out math problems on your own without knowing/having forgotten the intended solutions/techniques is so much more rewarding
I remember working on the generalised version of this problem. If we have a rectangle with sides x and y, then the number of rectangles is (sum of first x natural numbers)*(sum of first y natural numbers) i.e. [x(x+1) + y(y+1)]/2
huh? you keep adding consecutive perfect cubes, and each result is a perfect square? That's just weird.
codediporpal Yes, specifically they are the squares of the triangular numbers.
Take two consecutive triangular numbers Tn and Tn-1 and let's look at the difference of their squares. First note that a^2-b^2=(a-b)(a+b), so we have:
(Tn)^2 - (Tn-1)^2
= [(n(n+1) + n(n-1))/2]
• [(n(n+1) - n(n-1))/2]
The first factor in this product reduces to n^2 and the second factor reduces to n, thus the whole product is n^3.
So the solution to this puzzle is just (T8)^2 = 36^2 = 1296
codediporpal whoops didn't watch the end of the video...
You could also just square the nth triangular number. N being 8 in this case
I also noticed that when I saw the 1,3,6,10 emerging in my mind from the roots
I liked the way you solved using combination formula .
having dimentions of m x n, you can make the observation that rechtangle of size axb comes (m-1+a)(n-b+1)
then you summ if from a=1 to m and from b =1..n you will get summ((m-1+a) * n(n+1)/2 = m(m+1)/2 * n*(n+1)/2, substitute m = n = 8 to have your result
8,7,6...1. So... (8+1)×8/2 = 9×4 = 36.
That is one dimension.
36^2 = 1296
2 dimensions.
Done.
I used that same method
But about an 8 ×8×8 cube... I guess you can use the same formula to find .. 9c2 9c2 9c2 so 36^3 7 46656
For x by y rectangles, there are
((x+1)C2)((y+1)C2) rectangles.
The patteren i went for is 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3
Basiccally, number of rectangles in 1x1, then 2x2 - already counted, then 3x3...
Well as i continued watching, that was in fact the 1st solution he presented :D
Saved so much of my time, thank you so much ❤️
I came to (1+2+..+8)^2 like this: consider a single row of length 8. There are 8 possible lengths of rectangles. Namely those of length 1 through 8. In how many positions can length 1 be? well 8. And 2? well, one less so 7. And 8 length? well just in 1 position. So the answer for a 8*1 grid is 1+2+3+4+5+6+7+8. Now each of those possible rectangles could be 'streched' in the other dimension to any length up to 8 and positioned on the other rows. In how many ways can this happen with 8 rows? Well the same number of times. So we have 1+..+8 possible different rectangles on a row and 1+..+8 possible different configurations of those rectangles when considering the other rows. So the total number of possibilities is those to numbers multiplied. You end up with (1+..+8)^2.
Yeah... Looked at it and it was immediately obvious to me as well. Analytical minds 👌
Easier method:-
No. Of vertical lines:9
No. Of horizontal lines:9
Ans: 9C2 × 9C2
(To make a rectangle you need two vertical lines and two horizontal lines, so out of 9 lines lines you need two)
In 8×8 matrix,all combinations( like 1×2 and 2×1)resultant by adding gives 1296
It's simple. We see 9 horizontal lines and 9 vertical lines. Choosing 2 horizontal lines and 2 vertical lines will give us a rectangle. Therefore the answer is (9 C 2) ^ 2 = 36² = 1296
even smarter than my solution
I know I'm a bit late but you should do how many 3d rectangles are in a square shaped cube.
1x1x1
2×2×2
3x3x3
And so on
The answer is 46656. (9 choose 2)^3 = 46656.
(9C2)(9C2) rectangles ...
Simple math select two horizontal parellel lines and two vertical parallel to make a rectangle
Ganesh Prasad yeah he did that in the end of the video
@@panobato5984 oh I didn't watch till the end... I saw him counting and I got bored
Nice explanation .Thank you so much.
Please do more discrete math counting it’s so fun
If you are JEE aspirant, you know that this question is one of the beginner's solved example
By the way answer is 9C2*9C2
Because there are 9 lines, we select any 2 of them
Yes I easily figured out the answer
Answer: 0
You drew all the lines by hand, so none of them are perfectly straight, making non-right angles etc.
Jk, very nice problem and explanations :).
Btw, if one wants to count the number of squares, the answer is 1^2+2^2+3^2+...+n^2. The sum of squares, how cool is that!
n(n+1)(2n+1)/6
The white board tho
@@warioseggs that man is spitting fax
I solved it before I watched the video and got 1296.
I got to my answer by calculating each type of rectangle individually (1x1, 2x1, 2x2). However, you can simplify it further. you can realize that for a rectangle x1 is equal to x and by 2 is equal to x-1 and so on. For example the number of 1x2 rectangles is x(x-1) and the number of 3x4 is (x-2)(x-3).
Once you figure this out you have to find all the types of rectangles involved in the final shape so we can use (a+b+c+d+e+f+g+h)^2. We combine this with a = x, b = x -1... h= x - 7 and you get 64x^2-448x+784 (the formula ony works for x=8 but you can modify the math done simply). Plug in x=8 and you get 1296
I like the way he teaches. Man i miss learning math
Me: oh, 65.
This guy: heh, no.
wowitsWyatt 65??????
64*
@@blast-798 including the big one that would be the outside
@@ggtiket ...
so If the number of squares is "infinity x infinity" then the number of rectangles is
(1+2+3+4+5+........)^2 = (-1/12)^2 = 1/144 :D :D :D :D :D !!
No, 1+2+3+4+.... diverges and is not equal to anything.
@@DrBlueCow ...it's a joke
@@MyVash12349 ans. Is joke not the series!!
No, 1+2+3+4+5.. Diverges. İt is only true in number theory p-adic numbers
I did it in my own way
First I count the number of rectangle in the top row which is equal to n(n+1)/2
Since it is a square so it's left most side will also have n(n+1)/2
Then multiply both of these
You will get (n(n+1)/2)²
So I'm actually learning English and math at once. Thanks for your job, keep it up 🙂
0:13
problem: exists
my brain: it's easy, but let's try to make it easier
2hrs later: IT'S NOT EASY ENOUGH! SIMPLIFY!!!
Your enthusiasm never fails to make me smile!
Each rectangle has to have a whole number for height. There are fewer possible locations for taller rectangles. There are 8 spots for 1-high rectangles, 7 spots for 2-high rectangles, and so on. There are therefore 36 (the eighth triangular number) possibilities for height and vertical location combined, since they aren't independent. But height and width are independent for rectangles, so we square 36 to get all possible rectangles.
First I defined R(m, n) to be the number of rectangles in a (m, n) grid. Then I proved by induction that R(1, n) = n*(n+1)/2. Then I proved by induction again that R(m, n) = m*(m+1)/2 * n*(n+1)/2. A LOT more work than the combinatorics solution!
Total Rectangles = (9C2)^2.
Here "C" donates "Combinations"
64 because its not a rectangle anymore if lines go through it
Can easily be done using permutations and combinations.
To make a rectangle you need 2 horizontal and 2 vertical lines.
For above example
Method:
Choose 2 lines out of 9 vertical lines , i.e. 9c2
Now choose 2 lines out of 9 horizontal lines , i.e.
9c2
So no of rectangles= 9c2*9c2 =1296
First, we count the number of rectangles with height and lenght 1. There will be 8*8 = 64
Second, we count the number of rectangles that have either height 1 or length 1, but not both. For each column, there will be 7 2-square rectangles, 6 3-square rectangles... 1 8-square rectangle. In total there are 28 possible rectangles. Since there are 8 columns, there will be 28*8=224 possible rectangles with lenght 1. SImilarly, there are also 224 rectangles with height 1.
Third, we notice that each rectangle can be defined by 2 of it's opposite corners. For the rectangles we have already counted (height or lenght =1) there is only one pair of opposite corners that define the rectangle. For the rest of the rectangles there are 2 ways of defining that triangle (2 pairs of opposite corners). Let x denote the number of rectangles we haven't counted yet, that is, rectangles with both length and height greater than 1. We have 64choose2 = 2x + 2*224 . This is because 64choose2 represents all of the rectangles, except for the 64 rectangles with length and height 1, and in addition the rectangles with height and length greater than 1 are counted twice. Solving for x gives x = 784
Finally, the total number of rectangles is equal to x + 2*224 + 64 = 784 + 448 + 64 = 1296
I can't stop laughing after seeing his mega-sized old-fashioned microphone!!!
As soon as I saw number 1296, I knew it was 6^4 because I have memorised all my index powers from 1-10 raised to 1-10.
Cameron Gray Bruh I can barely remember all the numbers 1-20 squared.
Well I know all the ones up to 10 by heart, but 10+ I have to think about it a bit to figure it out...
9^10 ?
Nejx Well I already new my times tables up to 20. I a,ready knew the squares and the cubes up to 10 as well. Since I was pretty good with converting numbers between different bases I found it quite easy to learn all of these tables
Minechaîne Antoinecraft Bruh I legit forgot this one. I remembered the start which was like 3B 486Mil, the I got the last 6 numbers confused 😤😱😭. That is one of the hardest ones to remember as well.
I bet 1 raised to 1-10 was real tough
Really informative video ,thanks 😊 👍