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The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior
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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
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19 Feb 2011, 11:57




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Re: How many triangles can be formed
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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
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25 Feb 2011, 10:55
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
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25 Feb 2011, 23:20
Thanks for the question and great explanations.
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Re: How many triangles can be formed
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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
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10 Sep 2013, 06:25



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Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior
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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+48when 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
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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
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09 Jan 2016, 06:51
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
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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
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04 Aug 2018, 21:42
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? Sent from my Redmi Note 3 using GMAT Club Forum mobile app
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Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior
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04 Aug 2018, 22:41
dvishal387 wrote: 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? Sent from my Redmi Note 3 using GMAT Club Forum mobile appThe 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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Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior
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04 Aug 2018, 23:45
vaibhav1221 wrote: dvishal387 wrote: 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? Sent from my Redmi Note 3 using GMAT Club Forum mobile appThe 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. Okay..thanks Sent from my Redmi Note 3 using GMAT Club Forum mobile app
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Re: The sides BC, CA, AB of triangle ABC have 3, 4, 5 interior &nbs
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