A probability problem by Syed Hamza Khalid

Count the number of triangles in the picture above.

Hint: The answer is not 4. There's a faster way than simply counting them all!


The answer is 10.

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13 solutions

Chew-Seong Cheong
Jul 16, 2017

Relevant wiki: Binomial Coefficient

We note that a triangle is formed with two points on the slanting side and the point at the left-hand bottom vertex.

Therefore, the problem is equivalent to finding the number of ways to choose two points out of five on the slanting side which is N = ( 5 2 ) = 5 × 4 2 × 1 = 10 N = \displaystyle {5 \choose 2} = \dfrac {5\times 4}{2 \times 1} = \boxed{10}

Notation: ( n k ) = n ! k ! ( n k ) ! \displaystyle {n \choose k} = \dfrac {n!}{k!(n-k)!} denotes the binomial coefficient .

Moderator note:

The technique of using combinations does often work in general, but specific problems often have special cases that need to be considered. You might consider the general method to be more of a application of the Rule of Product that accounts for duplicates with division (which is itself simply a generalization of combinations).

Suppose we want to count the number of rectangles on an 8 by 8 grid. We'll do this by picking two points that represent the opposite corners of a rectangle.

Since there are 9 × 9 = 81 9 \times 9 = 81 corner points, we can naively suppose ( 81 2 ) = 6480 \displaystyle {81 \choose 2} = 6480 different rectangles.

One issue, though, is we may have picked our corners on the same row and column! So really our first choice has 81 points, but our second needs to exclude not only the point picked but everything on the same row and column; this represents 8 points on the row, 8 points on the column, so 81 1 8 8 = 64 81 - 1 - 8 - 8 = 64 choices. This means we have 81 × 64 = 5184 81 \times 64 = 5184 ways of picking the two corners.

However, also note we have duplicated each rectangle 4 times. Accounting for that brings our count to 5184 / 4 = 1296 5184 / 4 = 1296 rectangles, which is the correct answer.

Can you explain how the 9 x 9 corner points when you are counting the number of rectangles on an 8 by 8 grid?

Marvalisa Payne - 3 years, 10 months ago

Does this work for every problem like this?

Deva Craig - 3 years, 10 months ago

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I think so. But in this problem there is only one layer of triangles but if you add more layers the formulation can become quite complicated. Post more problems like this then I can try again.

Chew-Seong Cheong - 3 years, 10 months ago

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What about this ?

Munem Shahriar - 3 years, 10 months ago

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@Munem Shahriar The answer is:

Every side of the big square gives 5 triangles, therefore, 4 sides give 5 × 4 = 20 5 \times 4 = \boxed{20} .

Chew-Seong Cheong - 3 years, 10 months ago

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@Chew-Seong Cheong Is this the same concept?

Munem Shahriar - 3 years, 10 months ago

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@Munem Shahriar Same concept. I can always say ( 4 1 ) × 5 = 20 {4 \choose 1} \times 5 = 20

Chew-Seong Cheong - 3 years, 10 months ago

hi, loved the method but could you please explain what the (5/2) means as well as how you got 5 * 4 and 2 * 1 Thanks

Parth G - 3 years, 10 months ago

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I have added a note.

Chew-Seong Cheong - 3 years, 10 months ago

Neat solution!

Caleb Fong - 3 years, 10 months ago

Cool solution! You linked my first thought with combinatorics: ( n 2 ) \binom{n}{2} is the n'th triangle number!

Andrew Lamoureux - 3 years, 10 months ago

The best solution here is to count the triangles manually, and that is the easiest way. Using other methods (formulas) are complicated.

A Former Brilliant Member - 3 years, 10 months ago

There is an easier way to count the rectangles with no fear of repetition: each rectangle is determined by two horizontal lines (lines are formed by edges of the small squares) to be chosen from the 9 you see, as well as two verticals. So the number of possibilities is 9C2 times 9C2 = 36 x 36 = 1296.

Marcus Bizony - 3 years, 9 months ago

Here the total points are 6 out of which 5 are collinear and to form a triangle we need 3 points so total no of triangles formed is 6C3 (combination=C) but it also include the triangles formed by collinear points which cannot form a triangle but a straight line so we have to subtract 5C3 from 6C3.

Rasana Maharjan - 3 years, 9 months ago
Syed Hamza Khalid
Jul 16, 2017

Small triangles = 4

Triangles made by combining 2 triangles = 3

Triangles made by combining 3 triangles = 2

The big triangle = 1

Sum = 4 + 3 + 2 + 1 = 10

Answer: 10 triangles

I thought the small triangles could be 2 triangles since the angle you see them, that is why counted more.

Armando Castolo - 3 years, 10 months ago

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Can you explain what you meant better? Especially "since the angle you see them"? It is very unclear to me what you mean

Doug Reiss - 3 years, 10 months ago
Mohammad Khaza
Jul 30, 2017
  1. there are 4 small triangles......................[from the left side--1, 2, 3, 4]

  2. there is a 1 big triangle......................... [combining all(1+2+3+4)]

3.there are 3 triangles................................ [combining two------from the left side -(1+2), (2+3), (3+4)]

4.there are 2 triangles................................. [combining three---from the left side -(1+2+3), (2+3+4)]

so, there are--4+1+3+2=10 triangles

I can count 11 each time , am I counting the middle pair twice please?

Steve Soper - 3 years, 10 months ago

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look carefully, then count the small triangles and then the biggest triangle.you have got already 5 triangles.

now, take 2 small triangles and count from left or right sides[as there are 4 small triangles you can count in this way---(1+2) ,(2+ 3) ,(3+4) but you can not count (1+3) or,(1+4) because that does not make any triangles]

now, take 3 small triangles and count in this same way.

at last take 4 small triangles together .that is the biggest triangles which you have taken before. so, there is no more than 10 triangles.

if it helps you then tell me ,please

Mohammad Khaza - 3 years, 10 months ago

fun...missed by 2....

Dale Landefeld - 3 years, 10 months ago

I kind of agree with you but wouldn't it just be easier to add up the one side and multiply it by two. That's what I did and I got the right answer.

kara wilson - 3 years, 10 months ago

I'm dull, wrong app

Mark Spencer - 3 years, 10 months ago

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why?bro ,why?

Mohammad Khaza - 3 years, 10 months ago

Everyone's a beginner at some point in their life. Keep working on it! ;)

Pi Han Goh - 3 years, 10 months ago

I think the answer is not 10 but seven since: there is a big triangle that's =1 :then 4 triangles inside the bigger =4 combining the two triangles from right =2 A total of seven triangles. Winnie Jelagat;Mpesa Foundation
Academy.

Winnie Jelagat - 3 years, 8 months ago

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you counted in the right way but forgot to count the 3 triangles combining two.

i mean, the big triangle........(you counted)

4 small triangles..............(you counted)

2 triangles from right or left combining three.......(you counted)

3 triangles from right or left combining two........(you didn't count)

if need any other query,let me help you.

Mohammad Khaza - 3 years, 8 months ago

Thank you for that; agree with you: Winnie Jelagat

Winnie Jelagat - 3 years, 8 months ago
Kynan Lo
Aug 3, 2017

Count them

But what if there are too many to count?

Agnishom Chattopadhyay - 3 years, 10 months ago
Beth Davies
Jul 31, 2017

The amount of small numbers is 4 and you can work out how many it is quickly using 4!

1+2+3+4=10

didn't understand.

Mohammad Khaza - 3 years, 10 months ago

remember that 4! is the product of 4,3,2,1 and not the sum 4 ! = 4 × 3 × 2 × 1 = 24 4! = 4 \times 3 \times 2 \times 1 = 24

Andrew Lamoureux - 3 years, 10 months ago

I didn't got it.. 4! Isn't 10 man.. its 4×3×2×1=24 😲

Devansh Sharma - 3 years, 10 months ago

Can you explain what you mean? What are these small numbers that you are talking about?

Agnishom Chattopadhyay - 3 years, 10 months ago
Sean Bayley
Jul 31, 2017

Chew-Seong's answer is better, but I think it can be solved recursively. (n is the number of points between the two vertices on the hypotenuse, n 0 n \geq 0 )

f ( 0 ) = 1 f(0) = 1

f ( n ) = f ( n 1 ) + n + 1 f(n) = f(n-1) + n + 1

cool, and non-recursively with: f ( n ) = n 2 + 3 n + 2 2 f(n) = \frac{n^2 + 3n + 2}{2} EDIT: I just learned from brilliant to use "explicit" instead of "non-recursive" :)

Andrew Lamoureux - 3 years, 10 months ago

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that is awesome

Sean Bayley - 3 years, 10 months ago

Fascinating!

Zach Abueg - 3 years, 10 months ago

Glad to see a generalisation. Maybe you could explain how you formed the recurrence?

Agnishom Chattopadhyay - 3 years, 10 months ago
Calvin Dominic
Aug 3, 2017

In the following diagram, a triangle can be formed from any 3 of the 6 vertices.So it is the number of ways of choosing 3 vertices from 6 vertices.

But since 5 points lie on the same line(a triangle can't be formed by 3 points which lie on he same line), we should subtract the number of ways of choosing 3 vertices from 5 of those vertices.

So the answer is 6C3 - 5C3 = 20 - 10 = 10

It turns out that ( 6 3 ) ( 5 3 ) = ( 5 2 ) {6 \choose 3} - {5 \choose 3} = {5 \choose 2} . Is it in general true that, ( n k ) ( n 1 k ) = ( n 1 k 1 ) {n \choose k} - {n-1 \choose k} = {n-1 \choose k-1} ?

Agnishom Chattopadhyay - 3 years, 10 months ago

@Agnishom Chattopadhyay It is true. https://en.wikipedia.org/wiki/Pascal%27s_rule

Romero mukkolath - 3 years, 9 months ago
Ethan Smoller
Aug 2, 2017

I don't have much of a solution, more of a remark: It's pretty neat that the number of triangles in that bigger triangulated triangle added up to 10, the fourth triangular number!

Neat observation indeed. Have you tried explaining this connection?

Agnishom Chattopadhyay - 3 years, 10 months ago
Andre Bourque
Aug 1, 2017

There are 6 relevant points in the figure, and 5 colinear points. Thus the number is 6C3 - 5C3 = 20 - 10 = 10

Sohail Rasool
Aug 1, 2017

Since we know that a triangle requires 3 points to form but it should contain one left corner bottom point so we need to only select 2 points out of 5.Thus total ways of selecting 2 points (as one point is already selected) = 5C2 = 10

Mandy Newsome
Jul 31, 2017

Five in red and 5 in orange

I think you wanted to attach a diagram.

Agnishom Chattopadhyay - 3 years, 10 months ago
Zenin Easa
Aug 4, 2017

There are 6 points out of which 5 points lie on the same line. So, the answer is 6C3 - 5C3.

I don't agree with the hint in this problem. The best way is to count the triangles.

Really? But what if there were 10 small triangles instead of 4 small triangles?

Agnishom Chattopadhyay - 3 years, 10 months ago

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