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The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior

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The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior [#permalink]

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New post 19 Feb 2011, 11:35
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The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points respectively on them. How many triangles can be formed using these points as vertices.

A. 200
B. 205
C. 400
D. 410
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Re: How many triangles can be formed [#permalink]

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New post 19 Feb 2011, 11:57
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medanova wrote:
The sides BC,CA,AB of triangle ABC have 3,4,5 interior points respectively on them. How many triangles can be formed using these points as vertices.
a.200
b.205
c.400
d.410


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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Re: How many triangles can be formed [#permalink]

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New post 19 Feb 2011, 12:01
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.
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Re: How many triangles can be formed [#permalink]

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New post 25 Feb 2011, 10:55
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medanova wrote:
The sides BC,CA,AB of triangle ABC have 3,4,5 interior points respectively on them. How many triangles can be formed using these points as vertices.
a.200
b.205
c.400
d.410


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.
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Re: How many triangles can be formed [#permalink]

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New post 25 Feb 2011, 23:20
Thanks for the question and great explanations.
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Re: How many triangles can be formed [#permalink]

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New post 26 Feb 2011, 08:34
Thanks also from me.
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Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior [#permalink]

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Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior [#permalink]

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New post 10 Sep 2013, 10:57
medanova wrote:
The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior points respectively on them. How many triangles can be formed using these points as vertices.

A. 200
B. 205
C. 400
D. 410

[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
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Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior [#permalink]

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New post 09 Jan 2016, 06:35
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!

Bunuel wrote:
Bumping for review and further discussion.
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Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior [#permalink]

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New post 09 Jan 2016, 06:51
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mestrec wrote:
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!

Bunuel wrote:
Bumping for review and further discussion.


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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Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior [#permalink]

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New post 06 Mar 2016, 19:56
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
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Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior [#permalink]

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New post 27 Mar 2017, 10:27
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Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior   [#permalink] 27 Mar 2017, 10:27
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