Total points: 3+4+5=12;

Any 3 points out of 12 but those which are collinear will form a triangle, so \(C^3_{12}-(C^3_{3}+C^3_{4}+C^3_{5})=220-(1+4+10)=205\).

Answer: B.
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Using 1 point from each side of the triangle -
3C1 * 4C1 * 5C1 = 3 * 4 * 5 = 60

Using 1 point on 1 side and 2 points on the other side -
(3C1 * 4C2) + (3C1 * 5C2) + (4C1 * 3C2) + (4C1 * 5C2) + (5C1 * 4C2) + (5C1 * 3C2) = 145

Adding these we get = 60 + 145 = 205.

I was a little bit confused at first - I was not sure if we could use 2 of the interior points and 1 of the vertices of the existing triangle ABC (as the 3rd vertice) to form a new triangle. I needed to look at the question this way - How many triangles can be formed using only these points as vertices.

Thank you ,

I initially included also the points A B C in the calculation but now I read the question again and found my mistake.

Thanks for the question and great explanations.
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Thanks also from me.

[1*3*4]*5=60- when all the three points are on different sides..
[1*3+1*6]*5 +[1*10+1*3]*4+[1*10+1*6]*3=45+52+48-when two points on one side
total=60+45+52+48=205

Bunuel, is there a easier way to do the math, instead of multiplying 12 x 11 x 10 x 9 x 8 x 7.....?

As far I understood your solution, I would need to do that at the C 3 12 combination.

Regards and have a great year!

Hi,
Iwillhelp you out on that ..
the Q basically says there are three lines with 3,4,5 points respectively..
lets find ways triangle can be formed..
triangle requires three points,..
1) 2 points from line containing 3 points and one from remaining 4+5 points=3C2*(4+5)=27..
2) 2 points from line containing 4 points and one from remaining 3+5 points=4C2*(3+5)=48..
3) 2 points from line containing 5 points and one from remaining 4+3 points=5C2*(4+3)=70..
4) 1 point from each line containing 3,4,5 points =3*4*5=60..
total 27+48+70+60=205..


also you do not have to do the entire calculation in 12C3..
12C3=12!/3!9!= 12*11*10*9!/3!9!=12*11*10/3*2...
so you see you just have to multiply 12,11,and 10..
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very nice question...tried to analyze it first..then did not how to solve it..went with chetan2u 's method..
we can have 1 point on from 3, 1 point from 4, 1 point from 5: 3C1*4C1*5C1 = 60
we can have 1 point from 3, and 2 points from 5 = 3C1*5C2 = 30
we can have 1 point from 3, and 2 points from 4 = 3C1*4C2 = 18
we can have 1 point from 4, and 2 points from 3 = 4C1*3C2 = 12
we can have 1 point from 4, and 2 points from 5 = 4C1*5C2 = 40
we can have 1 point from 5, and 2 points from 3 = 5C1*3C2 = 15
we can have 1 point from 5, and 2 points from 4 = 5C1*4C2 = 30

now total: 60+30+18+12+40+15+30 = 205

hi,

different ways..
1) EASIEST approach..
lets see how many total can be made if none of the points were colliner that is NO three were in one line = 12C3 = 12!/9!3! = 220...
Due to collinear, extra triangles considered above = 3C3 + 5C3 +4C3 = 1+10+4 = 15..
total = 220-15 = 205

2) SECOND method
a) ONE point from each.
b) two points from one side and third from any of two other sides
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To expand on Chetan4u's second method, each triangle can be formed with one point from each of the 3 lines AB (3 points), BC (4 points) and CA (5 points), or with two points from one line, and the third point from either of the other two lines.

The calculations look like this:
Scenario 1) One point from each line = \(3*4*5 = 60\)
Scenario 2) Two points from one line and one from the other two lines = \(3c2*9 + 4c2*8 + 5c2*7 = 27+48+70=145\)

(Quick explanation here in case I just lost some of you - Taking 2 points from the line AB, which has 3 points on it can be done in 3c2 ways - choosing two points from the 3 available on AB. Then we can choose the third point from any of the remaining points on the other two lines BC or CA, which have a total of 9 points among them. So the number of triangles that can be formed with two points on line AB is \(3c2*9 = 27\). Similarly for the other two lines.)

Add up the two scenarios: \(60+145 = 205\)

Answer: B
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Hi, Can someone please clear my doubt- If we are joining three points ..one point from each side of original triangle then 4 triangles formed. Why we dont consider those triangles as well?

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The triangles need to be formed from the interior points i.e. treating the interior points as vertices. This is mentioned in the question, just go back and have a look.
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Okay..thanks

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We substract (C^3_{3}+C^3_{4}+C^3_{5}) from C^3_{12} to exclude the triangles that would be possible by using the points from the same sides, correct?

We subtract those combinations of points which are on a straight line (collinear) and thus do not form a triangle.
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Bunuel

Why can't we just pick 1 vertice from one side and any 2 vertices from the other 2 sides?
Side 1 (with 5 points) : 5 * 2C7
Side 2 (with 4 points) : 4 * 2C8
Side 3 (with 3 points): 3 * 2C9

And then sum them all?

I cannot find the gap in my reasoning here.