I thought about it as such. If you want a triangle of thickness 1, there are 3 columns to choose; of thickness 2 you can either pick 1st and 2nd column or 2nd and 3rd, so 2 options; of thickness 1, you have just one option. That’s 1+2+3=6. Then you can have either 1, 2, 3, or 4 layers of the triangle included, so that’s 4*6 = 24.
I love that you brute force all the drawing on the board instead of just making ditto marks (“). It shows an attention to detail and completeness that is necessary in mathematics and education. ❤
@@bprpmathbasicsthere are 25 triangles Instead of overcomplicating it,there are 3 parts per row, so just do 3!,which equals 6, Since there are 4 rows,do 6 times 4,which equals 24. Then since the whole thing is a large triangle (which counts too) Do 24+1=25 There are 25 triangles overall. Just pointing it out,ur still the goat
@@faheem-p7qYou're double-counting the whole triangular grid. It is already accounted for in your multiplication, it's where the width is 3 and height is 4. He specifically points out where he accounted for the full picture being one triangle at 2:35
@faheem-p7q, please think about what you're saying, it's simple math, and if it is SOMEHOW really that hard, then why don't you just count every single one, cause if you do that you'll get 24. It's easy and simple, you can learn it in like 2nd grade
For the rectangles, each rectangle is defined by two horizontal lines and two vertical lines. There are 6C2 many combinations of 2 vertical lines, and 5C2 many combinations of 2 horizontal lines. 6C2=(6*5)/(2*1)=15 5C2=(5*4)/(2*1)=10 Multiply them together and we get the answer of 150 different rectangles.
I first started by the left and counting the number of ways I could choose a side with 1, 2, 3 or 4 grid sides (turned out to be 10), then I multiplied by 4C2 (assuming it is a square grid). I also thought about multiplying by 5C2 but clearly that’s wrong too, giving 100 instead of 150. What went wrong? Edit: I find that by only choosing those sides, I confine myself to only those lines and not the other vertical lines, therefore overlooking some cases. This was just totally wrong as I didn’t even make the horizontal sides for the rectangle.
A midway technique I used: Count only the top triangles - 3 singles, 2 doubles and 1 triple triangle = 6. Then multiply by 4 because there are 4 sections downwards. 6*4=24 I feel like I'd be more likely to make an error if I tried to count them all by missing something or if I tried to come up with something more complex like you did.
I did a hybrid for the triangle calculation. I counted all the 1-height triangles which was 6. Then I multiplied by 4 because each base just adds a similar triangle for each of the 1-layer triangles. Less tedious that trying to count them all, but also simpler and more intuitive than having to remember the choose function.
You could also define a rectangle as being a selection of 2 points not on the same column or row as each other. There are (5+1)*(4+1)=30 points to select from. Our next point has some limitations, however, it has to be in a different column and row. Since each column has 5 other points(not counting the one we already selected), and each row has 4 other points(not counting the point we selected), there are 9 total points plus the point we are on that we can not pick. Thus we cannot pick 10 of the 30 points as a second point. So we can pick 30-10=20 points. 30*20=600. However, we have overcounted by 2. We could select point A then B. Or point B then A. Thus, we divide by 2 to get 300. Finally, each rectangle has 2 pairs of corners we can define it with, so we divide by 2 again to get 300/2=150.
Thanks! These brainteasers always bugged me. Choosing edges was the insight I was missing. Applying it to rectangles: (6 choose 2) times (5 choose 2) = 150
This is how I did it. You can make 3 triangles each column and then extend it each row. Then 2 for each row by combining 2 of them. Then 1 for each by combining the entire column for each row. There are 4 rows so the answer would be: 4*3 + 4*2 + 4*1 = 24. It can also be written like this when working with larger rows and columns: rows * columns / 2 * (columns + 1)
Solution to 4:59 question: (6 choose 2) x (5 choose 2) = 150 rectangles And if you want to exclude squares (because people forget all squares are rectangle) then: All rectangles - Squares 150 - (20 + 12 + 6 + 2) = 110
I didn't get that method. It's kinda complicated to understand why it works. I believe you can just multiple the number of horizontal lines to the triangular number for the (number of vertical lines -1) In that case it's 4*(3+2+1) = 4*6 = 24.
For rectangles it's a bit different, you'll have to multiply the triangular number for (number of vertical lines - 1) to the triangular number of (number of horizontal lines - 1). So in the case of given question it'd be (5+4+3+2+1)*(4+3+2+1) = 15*10 = 150
Its not 24,even google says its 25 How i did this triangle question,i used this method Each row has 3 triangles that you can see,to get the total amount of triangle combinations,just do 3! Factorial,which equals 6 Since there are 4 rows,just do 6 times 4 which equals 24 Then the whole thing is just a big triangle,so do 24+1 which equals 25 There are 25 triangles overall.
@@faheem-p7q well, the guy literally counted them on video, it's 24. Your factorial method doesn't work. Well, it does if you don't add the last "big one" triangle which was already counted in the 24 triangles, but only because triangular number for 3 is also 6. Thing is: you have 3 1-section triangles. If you go down a section, you'll get 3 2-sections triangles. Then 3 3section triangles, then 3 4-section triangles. That's 3*4 triangles total. Then from that 3 1-section triangles at the top you can make 2 2-section triangles, going down you have 2 4-section triangles, then 2 6-section triangles, then 3 8-section triangles. That's 2*4 triangles total Then you have 1 3-section triangle at the top. 1 6-section triangle if you go down. 1 9-section triangle and 12-section triangle (the "big one") That's 1*4 triangles total. So you have: 3*4 + 2*4 + 1*4 = (3+2+1)*4 = 24
Triangle has 2 sides and 1 base so just count the number of bases. First look at rows: we have the peak/top level + top 2 levels + top 3 levels + full height = 4 rows of bases. Then the columns: we have 3 side to side + 2 overlaps + 1 big base = 6 columns of bases. Multiple those to get the answer: 4 rows × 6 columns = 24 units of triangles This works for the rectangle as well: 10 possible rows × 15 possible columns = 150 rectangles
Could it just be said as 4 triangles of different sizes, times the total number of triangles that can be made in each one? 6 triangles in each triangle, there are 4 of them, so 24. Am I wrong?
I appreciate but... Do 6 (top) x number of rows => 24 Or do it like: 3 single columns of 4 => 12 2 duo-column of 4 => 8 1 trio-column of 4 => 4 then 12+8+4= 24 Mixing both method is harder. 😅 Edit: oh ok you have an elegant solution too.
(6*5)/2*(5*4)/2 = 150 I know the parentheses dont matter but they look cleaner imo. There are 6 potential borders for the rectangle on one axis and 5 on the other, but the order in which theyre picked doesnt matter which is why both multiplications are divided by two (to not take into account the order in which the borders are chosen). Since two different borders must be chosen from each, you multiply the number of all available options with that number minus one to account for the initial border on the axis not being available anymore. As such for an n*m grid you can say that it contains n*(n-1)/2*m*(m-1)/2 rectangles
I might be wrong about that, but I think when you using the / symbol for division parenthesis do metter and the / symbol represent the horizontal line, not the ':'. So correct parentheses would be: (6*5/2)*(5*4/2) Cause in your version it's more like [(6*5):(2*5*4)]:2
Don't quite get the part about 4*3 / 2*1 but if we have n lines from the top vertise and m horizontal lines, can we generalize number of triangles like sum(1..n-1) * m ?
For the triangle, I thought “well only the top pieces can make a triangle given all other pieces are rectancles” so I counted the ways to make triangles for the top layer and then just said every straight line below those adds the same number to the total
Still not entirely sure how you get that "2*1" unless you mean to count two given sides only one time. You have 4 lines you can start with, and three you can end with, but you want to count any two given sides only one time, thus "(4*3)/(2*1). For the rectangle we have six vertical lines we can start with and five we can end with, and we want to count any two given sides only one time. (6*5) / (2*1) = 30 / 2 = 15 rectangles per row we have five horizontal lines we can start with and 4 we can end with, and we want to count any two given sides only one time. (5*4) / (2*0) = 20 / 2 = 10 rectangles per column We now take the number of rectangles per column multiply by the number of rectangles per row, but we also want to count any given rectangle only once. (15*10) / (2*1) = 150 / 2 = 75 total possible rectangles. No, this is wrong. I have 15 rectangles per row, and can make rows up to 10 varying heights. You want to calculate your number of possible vertical widths and multiply it by your total number of horizontal widths. ( (6 * 5) / (2 / 1) ) * ( (5 * 4) / ( 2 * 1) ) = ( 30 / 2 ) * ( 20 / 2) = 15 * 10 =150
Here's my bonus question: what are the number of quadrilaterals (four sided shapes) in the first image? Note: this follows from the other two questions.
-How many rectangles are... -Ha easy, there is no right angle so zero. None of these quadrilaters are rectangles. It's a trick question! -in this new shape? -Ha bummer...
@idkdamsel The entire triangles divisions? I got 24 basically by doing what BpRp did in the beginning, I traced the largest triangle first, and split into two, then into three, then I did that up the rungs until you get the smallest 6 triangles.... I can't trace any more after 24.
Without overlaps (min side) 1×n: 20+16×2+15+13+5 2×n: 17+12+10+3 3×n: 7+6+5 4×n: 2+1 A total of 148 rectangles To me, squares are rectangles with equal sides so those count to. [Used the Square,line,edge method to count, starting at 1×1s]
@@colinmaranan4709 Actually, your exactly right. The definition of a square is a shape with four equal sides and angles, but according to the definition of a rectangle, it is a shape that has four sides and four right angles... Even the definition of a square says "a rectangle with four equal sides."..... According to Merriam Websters dictionary.
I find the rectangles easier. There are ∆5 possible rectangles in the first row and in the first column. All other rectangles are a combination of those two. So you get ∆5*∆5 = 15*15 = 225. What helped me figure that out though was your drawing them separately. I dod that for the row and column in my head, and realized that it would map nicely to a 15x15 grid. Then I tested it it with a 2x2 and 3x3 square to check. You do indeed get 9 and 36 when you count.
Note: the ∆ here means the triangle of a number, which is defined as ∆n = 1+2+3+...+n = n(n+1)/2 Why it is called a triangle, I will leave you to discover (if you've never heard of it).
Just start with the biggest triangle there which is the whole diagram Then start from the top 3 Total now = 4 Top 3 as one Top 3 from left Top 3 from right Total = 7 Top 6 Total now = 10 Top 6 as one Top 6 from left Top 6 from right Total = 13 Top 9 Total now = 16 Top 9 as one Top 9 from left Top 9 from right Total now = 19 Top 12 Total now = 22 Top 12 as one Top 12 from left Top 12 from right Total now = 25 PS: as i now iterate over every possible combination possible i spotted an error in my initial observation The triangle itself as a whole can be split into 3 both containing the horizontal and vertical lines Making the total *28* instead of 27 When we add the three to the total we got before 25 And adding from the left and right for the whole thing as we did for every segment we get a *GRAND TOTAL* of 30 triangles possible…. In foresight I might still be missing some combos But there are 30 being observed right now Scratch that 🤦♂️ I already considered the top 12 in all it’s possibilities so the final 2 are duplicates It’s 28 in total That’s why we have formulae 😅 I guess To avoid human errors such as this
1. There are 4 rows. Look at the top, there are 6 possible triangles 6 x 4 = 24 2. There are three types of triangles: - triangle with no line from the vertex to the base - triangle with one line from the vertex to the base - triangle with two lines from the vertex to the base Look at the topmost row. Meaning, exclude talks about rows that lead us to multiply. At the topmost row, there are: - two triangles with one line - three triangles with no lines and only - one triangle with two lines Now, consider that there are four rows, multiply each of those values by 4. - 4(2) = 8 - 4(3) = 12 - 4(1) = 4 Add them together: 8 + 12 + 4 = 24 Some argue that there is 25, as we didn't account the biggest triangle apparently. However, it was already accounted for as it basically the two-lined triangle that was descended to the fourth and last row, therefore it was counted in. I hope you understand. For me honestly, I need more clarity with how you got 30.
For the 1 by 2 and 2 by 1, it's 31 total. 1 by 3 and 3 by 1, it's 22, 53 total. 1 by 4 and 4 by 1, it's 13, 66 total. 5 by 1, it's 4, 70 total. 2 by 3 and 3 by 2, it's 17, 87 total. 2 by 4 and 4 by 2, it's 10, 97 total. 5 by 2, it's 3, 100 total. 3 by 4 and 4 by 3, it's 7, 107 total. 5 by 3, it's 2, 109 total. 5 by 4, it's 1, 110 total. You need to count only the rectangles, not the squares. You'll see I'm right
I have seen this problem in one of those late night to early morning game shows in my country where they offer something like 5k for solving this problem yet no one ever solves it
What if there were 4 columns? The answer is the summation of every number below the columns count multiplied by the row count. In this case: (1+2+3)*4 You are mistaken because 1+2+3 = 3!
@@devdog7409 you're right. I thought it was n!, but it's Tn, where Tn is the nth Triangular number Tn = (n²+n)/2. So the correct closed from is t = r × Tc, t Triangles r Rows c Columns
In a vector graphics program, it would be since that keeps all the verticies, but here, they verticies dont matter since the lines remain straight and a mathmatical line has infinite points anyways.
Or for every row just do 3! Which equals 6,so 6 times 4 equals 24,then since the whole thing is just a big triangle, Do 24+1 which equals 25 There are 25 triangles overall.
But if each row has a total triangle combination of 6, 6 X 4=24 then the big triangle add that equals 25 Oh and before u even argue this point,search it on google,even google says it is 25
I dont like these kind of questions, because it can go from the least to the most, all answers are right and wrong, it all depends on who asks it. If they want to just count the top 3, and not count when you merge multiple shapes into a triangle, or if they want to count it by combining the shapes into triangles, its on them to pick the correct answer.
that was hella confusing lol...i just counted 6 at the top tier then multiplied by the 4 levels cuz you would just repeat the same pattern on each tier so 6x4 = 24........UNLESS its a 2D drawing of a pyramid...then there's only 16 triangles
I laughed a bit when i saw the question at the end🤣 I think we were tought that at 1st or second grade im not sure... but man, if you dont have an awnser, you know nothing about math😅
Before watching the video, I also did a counting method But instead of separating into 12, 8, 4 I noticed that all the triangles have the top apex in common, so we fix that choice. There are four levels, take the first level. Each level is split into 3. Either we can have triangles with equal split {1,1,1} 3 such triangles [base length '1 unit' each] or triangles with a split of {2,1} {1,2}, 2 such triangles ['base length of 2 units for one of the triangles and 1 unit for the other'] Or use the whole level {3} as the base, 1 such triangle. [whole base length of '3 units' for the big triangle on this level] I.e. this is (3 + 2 + 1) But there are four levels so (3+2+1)(4) = 24 and that's how I did it. Then I noticed 4 choose 2 from 3 + 2 + 1, but did not see the geometric connection until I saw the video. which also allowed me to see where 4 choose 1 geometrically came from in the picture.
I can see the other way to see 4 choose 2. Each level when 'split into 3 (smaller triangles)' has '4 separate points' on that level, and 2 of those points are chosen as the other vertices of the triangle. hence, 4 choose 2.
Before I know the answer mine is 16, now I will finish the video and see. Edit: so 24 is the total I should have looked a little longer lol. Rushing things can make you miss steps.
![ATEEZ(에이티즈) - [GOLDEN HOUR : Part.2] Preview](http://i.ytimg.com/vi/X4PfgRPBUVo/mqdefault.jpg)
Math for fun, how many rectangles? ruclips.net/video/Uq9OXC0Gzgw/видео.htmlsi=zDsPYEpR5MUknuom
after watching long enough, i got 25. the last 1 is coming from the whole triangle that's containing 24 triangles in it as you count
1x1 => 4*5 = 20
1x2 => 4*4 = 16
1x3 => 4*3 = 12
1x4 => 4*2 = 8
1x5 => 4*1 = 4
2x1 => 3*5 = 15
2x2 => 3*4 = 12
2x3 => 3*3 = 9
2x4 => 3*2 = 6
2x5 => 3*1 = 3
I'm sensing a patttern...
1+2+3+4+5=15
15+30+45+60=150
There are 150 rectangles.
24
110
@michaelbujaki2462 don't count the 1 by 1, 2 by 2, 3 by 3, or 4 by 4, it's not rectangles
"Do I have any other colors? Of course."
Never doubted you, dude.
lol thanks!
@@bprpmathbasics can you do equation of circle next? by finding radii and centre point coords?
Damn, I was so confused on all the comments mentioning 150, turns out it's the answer for different question.
I thought about it as such. If you want a triangle of thickness 1, there are 3 columns to choose; of thickness 2 you can either pick 1st and 2nd column or 2nd and 3rd, so 2 options; of thickness 1, you have just one option. That’s 1+2+3=6. Then you can have either 1, 2, 3, or 4 layers of the triangle included, so that’s 4*6 = 24.
At least 28.
I really hope that 28 is a joke
I love that you brute force all the drawing on the board instead of just making ditto marks (“). It shows an attention to detail and completeness that is necessary in mathematics and education. ❤
Thank you!
@@bprpmathbasicsthere are 25 triangles
Instead of overcomplicating it,there are 3 parts per row, so just do 3!,which equals 6,
Since there are 4 rows,do 6 times 4,which equals 24.
Then since the whole thing is a large triangle (which counts too)
Do 24+1=25
There are 25 triangles overall.
Just pointing it out,ur still the goat
@@faheem-p7qYou're double-counting the whole triangular grid. It is already accounted for in your multiplication, it's where the width is 3 and height is 4.
He specifically points out where he accounted for the full picture being one triangle at 2:35
@@faheem-p7qbro u dumb or what😂
@faheem-p7q, please think about what you're saying, it's simple math, and if it is SOMEHOW really that hard, then why don't you just count every single one, cause if you do that you'll get 24. It's easy and simple, you can learn it in like 2nd grade
For the rectangles, each rectangle is defined by two horizontal lines and two vertical lines.
There are 6C2 many combinations of 2 vertical lines, and 5C2 many combinations of 2 horizontal lines.
6C2=(6*5)/(2*1)=15
5C2=(5*4)/(2*1)=10
Multiply them together and we get the answer of 150 different rectangles.
Triangles have 3 sides, don't count the quadralaterals (4 sides) as triangles.
@@richardhole8429 I'm talking about the rectangle problem presented at the end.
how many squares?
I first started by the left and counting the number of ways I could choose a side with 1, 2, 3 or 4 grid sides (turned out to be 10), then I multiplied by 4C2 (assuming it is a square grid). I also thought about multiplying by 5C2 but clearly that’s wrong too, giving 100 instead of 150. What went wrong?
Edit: I find that by only choosing those sides, I confine myself to only those lines and not the other vertical lines, therefore overlooking some cases. This was just totally wrong as I didn’t even make the horizontal sides for the rectangle.
@@richardhole8429 Can't even spell "quadrilaterals" right.
A midway technique I used:
Count only the top triangles - 3 singles, 2 doubles and 1 triple triangle = 6.
Then multiply by 4 because there are 4 sections downwards.
6*4=24
I feel like I'd be more likely to make an error if I tried to count them all by missing something or if I tried to come up with something more complex like you did.
did the same :-)
I know it's 4*(3+2+1). You count the horizontal areas and multiply by the sumatory of the vertical areas.
I did a hybrid for the triangle calculation. I counted all the 1-height triangles which was 6. Then I multiplied by 4 because each base just adds a similar triangle for each of the 1-layer triangles. Less tedious that trying to count them all, but also simpler and more intuitive than having to remember the choose function.
Oh,you got it right, this is easier 😊 thanks a lot
That’s what I did too. Yay for us.
You could also define a rectangle as being a selection of 2 points not on the same column or row as each other. There are (5+1)*(4+1)=30 points to select from. Our next point has some limitations, however, it has to be in a different column and row. Since each column has 5 other points(not counting the one we already selected), and each row has 4 other points(not counting the point we selected), there are 9 total points plus the point we are on that we can not pick. Thus we cannot pick 10 of the 30 points as a second point. So we can pick 30-10=20 points. 30*20=600. However, we have overcounted by 2. We could select point A then B. Or point B then A. Thus, we divide by 2 to get 300. Finally, each rectangle has 2 pairs of corners we can define it with, so we divide by 2 again to get 300/2=150.
Indeed Is a (6*(6-1))*(5*(5-1))/4. It applies to all rectangles that equation btw, and the 4 dividing is a constant.
Thanks! These brainteasers always bugged me. Choosing edges was the insight I was missing.
Applying it to rectangles: (6 choose 2) times (5 choose 2) = 150
This is how I did it. You can make 3 triangles each column and then extend it each row. Then 2 for each row by combining 2 of them. Then 1 for each by combining the entire column for each row. There are 4 rows so the answer would be:
4*3 + 4*2 + 4*1 = 24.
It can also be written like this when working with larger rows and columns:
rows * columns / 2 * (columns + 1)
Solution to 4:59 question:
(6 choose 2) x (5 choose 2) = 150 rectangles
And if you want to exclude squares (because people forget all squares are rectangle) then:
All rectangles - Squares
150 - (20 + 12 + 6 + 2) = 110
I didn't get that method.
It's kinda complicated to understand why it works.
I believe you can just multiple the number of horizontal lines to the triangular number for the (number of vertical lines -1)
In that case it's 4*(3+2+1) = 4*6 = 24.
For rectangles it's a bit different, you'll have to multiply the triangular number for (number of vertical lines - 1) to the triangular number of (number of horizontal lines - 1).
So in the case of given question it'd be (5+4+3+2+1)*(4+3+2+1) = 15*10 = 150
Its not 24,even google says its 25
How i did this triangle question,i used this method
Each row has 3 triangles that you can see,to get the total amount of triangle combinations,just do 3! Factorial,which equals 6
Since there are 4 rows,just do 6 times 4 which equals 24
Then the whole thing is just a big triangle,so do 24+1 which equals 25
There are 25 triangles overall.
@@faheem-p7q well, the guy literally counted them on video, it's 24.
Your factorial method doesn't work. Well, it does if you don't add the last "big one" triangle which was already counted in the 24 triangles, but only because triangular number for 3 is also 6.
Thing is: you have 3 1-section triangles. If you go down a section, you'll get 3 2-sections triangles. Then 3 3section triangles, then 3 4-section triangles.
That's 3*4 triangles total.
Then from that 3 1-section triangles at the top you can make 2 2-section triangles, going down you have 2 4-section triangles, then 2 6-section triangles, then 3 8-section triangles.
That's 2*4 triangles total
Then you have 1 3-section triangle at the top. 1 6-section triangle if you go down. 1 9-section triangle and 12-section triangle (the "big one")
That's 1*4 triangles total.
So you have: 3*4 + 2*4 + 1*4 = (3+2+1)*4 = 24
@@faheem-p7q lmfao bro the whole thing is literally counted, it's the last base, which is counted
When you do the last row with all 3 that is the whole triangle.
24 (3+2+1)×4=24 triangles.
(4+3+2+1)(5+4+3+2+1)
(10)(15)
150 rectangles
Triangle has 2 sides and 1 base so just count the number of bases. First look at rows: we have the peak/top level + top 2 levels + top 3 levels + full height = 4 rows of bases. Then the columns: we have 3 side to side + 2 overlaps + 1 big base = 6 columns of bases. Multiple those to get the answer: 4 rows × 6 columns = 24 units of triangles
This works for the rectangle as well: 10 possible rows × 15 possible columns = 150 rectangles
Could it just be said as 4 triangles of different sizes, times the total number of triangles that can be made in each one? 6 triangles in each triangle, there are 4 of them, so 24. Am I wrong?
Sounds good to me.
Just do it the other way around. it is 6 per level. (3 single, 2 double, 1 overall) and multiply it times the four levels. No difficult math needed.
I appreciate but...
Do 6 (top) x number of rows => 24
Or do it like:
3 single columns of 4 => 12
2 duo-column of 4 => 8
1 trio-column of 4 => 4 then 12+8+4= 24
Mixing both method is harder. 😅
Edit: oh ok you have an elegant solution too.
5*6*5*4/2/2= 150
15 ❤
(6*5)/2*(5*4)/2 = 150
I know the parentheses dont matter but they look cleaner imo.
There are 6 potential borders for the rectangle on one axis and 5 on the other, but the order in which theyre picked doesnt matter which is why both multiplications are divided by two (to not take into account the order in which the borders are chosen).
Since two different borders must be chosen from each, you multiply the number of all available options with that number minus one to account for the initial border on the axis not being available anymore.
As such for an n*m grid you can say that it contains n*(n-1)/2*m*(m-1)/2 rectangles
I might be wrong about that, but I think when you using the / symbol for division parenthesis do metter and the / symbol represent the horizontal line, not the ':'.
So correct parentheses would be:
(6*5/2)*(5*4/2)
Cause in your version it's more like [(6*5):(2*5*4)]:2
Don't quite get the part about 4*3 / 2*1 but if we have n lines from the top vertise and m horizontal lines, can we generalize number of triangles like sum(1..n-1) * m ?
We choose 2 horizontal lines and 2 vertical lines.
So, number of rectangles
= 5C2 × 6C2
= 10 × 15
= 150
Why are you counting quadralaterals instead of triangles. You are answering a different question. By the way there are zero rectangles.
@@richardhole8429If you watch to the end, there's a bonus question, which is indeed about rectangles.
@@richardhole8429 right 😂
For the triangle, I thought “well only the top pieces can make a triangle given all other pieces are rectancles” so I counted the ways to make triangles for the top layer and then just said every straight line below those adds the same number to the total
basically you have 4*6 made of of 3*1 2*2 and 1*3 and can be extended by 4 rows, giving 24 in total.
Still not entirely sure how you get that "2*1" unless you mean to count two given sides only one time. You have 4 lines you can start with, and three you can end with, but you want to count any two given sides only one time, thus "(4*3)/(2*1).
For the rectangle we have six vertical lines we can start with and five we can end with, and we want to count any two given sides only one time.
(6*5) / (2*1) = 30 / 2 = 15 rectangles per row
we have five horizontal lines we can start with and 4 we can end with, and we want to count any two given sides only one time.
(5*4) / (2*0) = 20 / 2 = 10 rectangles per column
We now take the number of rectangles per column multiply by the number of rectangles per row, but we also want to count any given rectangle only once.
(15*10) / (2*1) = 150 / 2 = 75 total possible rectangles. No, this is wrong.
I have 15 rectangles per row, and can make rows up to 10 varying heights. You want to calculate your number of possible vertical widths and multiply it by your total number of horizontal widths.
( (6 * 5) / (2 / 1) ) * ( (5 * 4) / ( 2 * 1) )
= ( 30 / 2 ) * ( 20 / 2)
= 15 * 10
=150
I made 4(levels) * 6(variations)
That's the way I did it, very quick method.
Here's my bonus question: what are the number of quadrilaterals (four sided shapes) in the first image?
Note: this follows from the other two questions.
-How many rectangles are...
-Ha easy, there is no right angle so zero. None of these quadrilaters are rectangles. It's a trick question!
-in this new shape?
-Ha bummer...
24
0:04 I'm currently counting 24. I initially counted 12, then 14.
Sure sure
27
@@ajayimoses2495 Interesting. How did you get 27?
@@ajayimoses2495yhyh
@idkdamsel The entire triangles divisions? I got 24 basically by doing what BpRp did in the beginning, I traced the largest triangle first, and split into two, then into three, then I did that up the rungs until you get the smallest 6 triangles.... I can't trace any more after 24.
Before watching:
It has to have the top point as 1 vertice. The other 2 points are in the same line.
6 for each line, 24
5:12 Ooo. Tricky. Assuming all four sided shapes are rectangular, 20+5+4+...... Nevermind. It's a lot of rectangles. I'm too tired for this one.
Without overlaps (min side)
1×n: 20+16×2+15+13+5
2×n: 17+12+10+3
3×n: 7+6+5
4×n: 2+1
A total of 148 rectangles
To me, squares are rectangles with equal sides so those count to.
[Used the Square,line,edge method to count, starting at 1×1s]
@@colinmaranan4709 Actually, your exactly right. The definition of a square is a shape with four equal sides and angles, but according to the definition of a rectangle, it is a shape that has four sides and four right angles... Even the definition of a square says "a rectangle with four equal sides."..... According to Merriam Websters dictionary.
@@colinmaranan4709 I was having a hard time because I wasn't counting the squares.
I find the rectangles easier. There are ∆5 possible rectangles in the first row and in the first column. All other rectangles are a combination of those two. So you get ∆5*∆5 = 15*15 = 225.
What helped me figure that out though was your drawing them separately. I dod that for the row and column in my head, and realized that it would map nicely to a 15x15 grid.
Then I tested it it with a 2x2 and 3x3 square to check. You do indeed get 9 and 36 when you count.
Note: the ∆ here means the triangle of a number, which is defined as
∆n = 1+2+3+...+n = n(n+1)/2
Why it is called a triangle, I will leave you to discover (if you've never heard of it).
You should notice, that the main rectangle is not 5x5, but 5x4, so it'll be ∆5*∆4 = 15*10 = 150.
@@F1r1at Ah, thank you.
I am happy to know that counting them worked at least.
I counted the first row: 1 big, 2 middle, 3 small triangles = 6 triangles per row
4 rows makes 24 triangles
One hundred and ten (if you include oblongs).
There're also forty squares of various sizes.
I hope the answer helps you with your future videos.
Just start with the biggest triangle there which is the whole diagram
Then start from the top 3
Total now = 4
Top 3 as one
Top 3 from left
Top 3 from right
Total = 7
Top 6
Total now = 10
Top 6 as one
Top 6 from left
Top 6 from right
Total = 13
Top 9
Total now = 16
Top 9 as one
Top 9 from left
Top 9 from right
Total now = 19
Top 12
Total now = 22
Top 12 as one
Top 12 from left
Top 12 from right
Total now = 25
PS: as i now iterate over every possible combination possible i spotted an error in my initial observation
The triangle itself as a whole can be split into 3 both containing the horizontal and vertical lines
Making the total *28* instead of 27
When we add the three to the total we got before 25
And adding from the left and right for the whole thing as we did for every segment we get a
*GRAND TOTAL* of 30 triangles possible….
In foresight I might still be missing some combos
But there are 30 being observed right now
Scratch that 🤦♂️
I already considered the top 12 in all it’s possibilities so the final 2 are duplicates
It’s 28 in total
That’s why we have formulae 😅 I guess
To avoid human errors such as this
1. There are 4 rows.
Look at the top, there are 6 possible triangles
6 x 4 = 24
2. There are three types of triangles:
- triangle with no line from the vertex to the base
- triangle with one line from the vertex to the base
- triangle with two lines from the vertex to the base
Look at the topmost row. Meaning, exclude talks about rows that lead us to multiply.
At the topmost row, there are:
- two triangles with one line
- three triangles with no lines
and only
- one triangle with two lines
Now, consider that there are four rows, multiply each of those values by 4.
- 4(2) = 8
- 4(3) = 12
- 4(1) = 4
Add them together:
8 + 12 + 4 = 24
Some argue that there is 25, as we didn't account the biggest triangle apparently. However, it was already accounted for as it basically the two-lined triangle that was descended to the fourth and last row, therefore it was counted in.
I hope you understand.
For me honestly, I need more clarity with how you got 30.
@@khairilhanafirosli9672
No no
Please read the full comment…
30 was an error, I got 28
@@ajayimoses2495 Then can you reiterate how you got 28?
For the 1 by 2 and 2 by 1, it's 31 total. 1 by 3 and 3 by 1, it's 22, 53 total. 1 by 4 and 4 by 1, it's 13, 66 total. 5 by 1, it's 4, 70 total. 2 by 3 and 3 by 2, it's 17, 87 total. 2 by 4 and 4 by 2, it's 10, 97 total. 5 by 2, it's 3, 100 total. 3 by 4 and 4 by 3, it's 7, 107 total. 5 by 3, it's 2, 109 total. 5 by 4, it's 1, 110 total. You need to count only the rectangles, not the squares. You'll see I'm right
I have seen this problem in one of those late night to early morning game shows in my country where they offer something like 5k for solving this problem yet no one ever solves it
3! × 4 = 24
3! for the "columns", 4 for the "rows"
What if there were 4 columns?
The answer is the summation of every number below the columns count multiplied by the row count. In this case:
(1+2+3)*4
You are mistaken because 1+2+3 = 3!
@@devdog7409 you're right. I thought it was n!, but it's Tn, where Tn is the nth Triangular number Tn = (n²+n)/2.
So the correct closed from is t = r × Tc,
t Triangles
r Rows
c Columns
I don't understand the 4x3 part? There are 4 lines but then it seems he chooses a line, 3. That can't be right? What am I missing?
What about the outer perimeter of the triangle...shouldn't that count for a separate triangle giving a total of 25?
It's the last green triangle he draw
A trillion billion and sixty nine thousand and one point zero triangles? Approximately?
Man, maths is so cool
Can you explain why the largest triangle is not a shape with 11 points and 11 sides?
2:52 - it's the bottom green triangle on the right
Having interior lines does not mean the overall shape is not a triangle.
Angles of 180° keep you on the same line. So the overall shape has just 3 outer lines.
In a vector graphics program, it would be since that keeps all the verticies, but here, they verticies dont matter since the lines remain straight and a mathmatical line has infinite points anyways.
At least 18.
I like how he says the answer at the start, but the pattern is that there is 6 in every level of it
I finally got one.
Or for every row just do 3! Which equals 6,so 6 times 4 equals 24,then since the whole thing is just a big triangle,
Do 24+1 which equals 25
There are 25 triangles overall.
No.. There are 24 triangles. You are counting the big triangle twice
But if each row has a total triangle combination of 6, 6 X 4=24 then the big triangle add that equals 25
Oh and before u even argue this point,search it on google,even google says it is 25
@@faheem-p7qyoure counting the big one twice, its already included in the 6 per row of the bottom row
Cuz Google is always right 😂😂😂@@faheem-p7q
(6 choose 2)*(5 choose 2) for the rectangle problem
I dont like these kind of questions, because it can go from the least to the most, all answers are right and wrong, it all depends on who asks it.
If they want to just count the top 3, and not count when you merge multiple shapes into a triangle, or if they want to count it by combining the shapes into triangles, its on them to pick the correct answer.
4!
The basic triangle (top to bottom) has 6 combinations. Each of those has 4 options. So 24.
Divide by 2 * 1 because isn't much of an explanation. This makes sense if you already understand it.
6 triangles per horizontal line was how I did it.
"I'm going to help students with advanced calculus problems!"
*clicks on Reddit*
"I may have been too ambitious..."
(4+3+2+1) x (5+4+3+2+1) aka possibile square in a column time possible square in a row. Is faster then counting and easier to calculate the 6C2x5C1
24 yupp
that was hella confusing lol...i just counted 6 at the top tier then multiplied by the 4 levels cuz you would just repeat the same pattern on each tier so 6x4 = 24........UNLESS its a 2D drawing of a pyramid...then there's only 16 triangles
6x4 is already smart for me
I don't get how 4 over 2 times 4 over 1 equals 24?
The first top section has 6 triangles. So then there are 4 levels... So 6x4=24. Done. 😊
All of them together make one big triangle, so shouldn't the answer be 25?
The top green one is the whole thing he did count it
@@edog9991Ah, it's the bottom green one actually with all 4 levels, but you're right it was counted. 👍
It should be 25 yes you can do 6*4+1
There's 6 every level, so it's 24. It's very easy if you think about it
@@jakobpedersen6317 you’re forgetting the overall thing is a triangle
I laughed a bit when i saw the question at the end🤣
I think we were tought that at 1st or second grade im not sure... but man, if you dont have an awnser, you know nothing about math😅
28
(6 choose 2) times (5 choose 2.) = 150
Can somebody tell me how many squares are there in 4by4 grid
Base 1: 4*4 = 16
Base 2: 3*3 = 9
Base 3: 2*2 = 4
Base 4: 1*1 = 1
Total = 30
@@willemakkermans4067 thank you
7
Before watching the video, I also did a counting method
But instead of separating into 12, 8, 4
I noticed that all the triangles have the top apex in common, so we fix that choice.
There are four levels, take the first level.
Each level is split into 3.
Either we can have triangles with equal split {1,1,1} 3 such triangles [base length '1 unit' each]
or triangles with a split of {2,1} {1,2}, 2 such triangles ['base length of 2 units for one of the triangles and 1 unit for the other']
Or use the whole level {3} as the base, 1 such triangle. [whole base length of '3 units' for the big triangle on this level]
I.e. this is (3 + 2 + 1)
But there are four levels
so (3+2+1)(4) = 24
and that's how I did it.
Then I noticed 4 choose 2 from 3 + 2 + 1, but did not see the geometric connection until I saw the video.
which also allowed me to see where 4 choose 1 geometrically came from in the picture.
I can see the other way to see 4 choose 2. Each level when 'split into 3 (smaller triangles)' has '4 separate points' on that level, and 2 of those points are chosen as the other vertices of the triangle. hence, 4 choose 2.
4!×3!= 144
150 rectangles
I came up with 25 triangles because the whole image together is also a triangle
Enough. The image has enough triangles.
148
I counted 7
25 is correct........... 24 as explained plus the main one
7
awwww. I missed a bunch. I was trying to be quick
4 rows times 6 column triangles = 24, that's my final solution before watching the full video.
I'd say 0. In the perspective of art, these are just rectangles affected by an optical illusion. Hahaha
Before I know the answer mine is 16, now I will finish the video and see. Edit: so 24 is the total I should have looked a little longer lol. Rushing things can make you miss steps.
25?
24
there are 3 triangles and 12 trapezoids lol
44
I get 24.
I got 22 by just counting.
I saw the thumbnail, counted, and without reading any comments, I came to 24. Dunno if I'm right. I'll watch the video now and see.
Wouldn’t it be 25, for the whole image being a triangle?
It's the last green triangle he draw
I counted 7. Edit: now I see there’s 16 😅
um i got 30??
I believe it's 21. Not sure how to actually do this correctly. EDIT: Dang I missed 3 of the green ones.
It's 150, right?
15 triangles?
Edit: 😂 wow i was so off
Hey, i got one right.
The whole triangle is another, so 25
That the 24th triangle.
@Apsolon 😂 haha I'm so dumb. I started thinking about how he missed it before I finished thinking about how it was clearly counted. Haha thank you
the entire shape is the 25th triangle, no?
The answer is 31.
6 times how many rows. so 6 * 4 is 24
The squares question i got 150. Is it?😊
It is for rectangles. If the question was how many squares, it would only be 40.
@@willemakkermans4067 alright,got it