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Number of Triangles

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by newton9 » Tue Mar 30, 2010 12:49 pm
7 points are marked on a straight line and another 8 points are marked on a second straight line with no points in common. How many triangles can be constructed with vertices from among the above points?


A. 91


B. 105


C. 196


D. 21


E. 14

I got 364 as answer. Based on 7C2 * 8 + 8C2 * 7

This question is from one of the other forums.

Did I miss anything ? Any thoughts?
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by harshavardhanc » Tue Mar 30, 2010 12:58 pm
newton9 wrote:7 points are marked on a straight line and another 8 points are marked on a second straight line with no points in common. How many triangles can be constructed with vertices from among the above points?


A. 91
B. 105
C. 196
D. 21
E. 14

I got 364 as answer. Based on 7C2 * 8 + 8C2 * 7

This question is from one of the other forums.

Did I miss anything ? Any thoughts?
I think you are right. i tried to do it by two ways :

method 1 :

Select 3 points out of total 15. Subtract number of ways of selecting 3 out of 7 collinear points and 3 out of 8 collinear points.

15C3 - 8C3 - 7C3 = 364.

Method 2 :

Select one point from 7 and two from other 8 collinear points OR one from 8 collinear and two from other 7 collinear .

8C1 * 7C2 + 7C2 * 8C1 = 364


I think, the options given there in forum are wrong. :)

any other takers?
Regards,
Harsha
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by ajith » Tue Mar 30, 2010 10:29 pm
newton9 wrote:7 points are marked on a straight line and another 8 points are marked on a second straight line with no points in common. How many triangles can be constructed with vertices from among the above points?


A. 91


B. 105


C. 196


D. 21


E. 14

I got 364 as answer. Based on 7C2 * 8 + 8C2 * 7

This question is from one of the other forums.

Did I miss anything ? Any thoughts?
You did not miss a thing, just a case of wrong options

https://www.beatthegmat.com/how-many-uni ... 51280.html
Always borrow money from a pessimist, he doesn't expect to be paid back.
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by eduardo_cruza » Tue Aug 30, 2011 7:15 pm
8C2*7C1 = 196
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