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by maihuna » Mon Aug 10, 2009 8:22 am
There are 70 points in a plane, no three of which are in the same straight line with the exception of 20 points, which are all in same straight line; find the number of triangles joining these straight lines.

Ans to follow, if possible give a general formula or little detail
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by real2008 » Mon Aug 10, 2009 9:55 am
i believe it to be 70C3-20C3
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by maihuna » Mon Aug 10, 2009 10:07 am
real2008 wrote:i believe it to be 70C3-20C3
Far far away, I have particularly tweaked this question with artificial numbers as many are just posting some ans and no logic...anyway your ans is far far away....
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by real2008 » Mon Aug 10, 2009 10:18 am
maihuna wrote:
real2008 wrote:i believe it to be 70C3-20C3
Far far away, I have particularly tweaked this question with artificial numbers as many are just posting some ans and no logic...anyway your ans is far far away....
then post your logic.......
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by kaulnikhil » Tue Aug 11, 2009 8:52 am
Dude is it 50c2*20c1???
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by xcusemeplz2009 » Tue Aug 11, 2009 10:27 am
9500
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by winnerhere » Wed Aug 12, 2009 5:29 am
"find the number of triangles joining these straight lines"????

what do u mean by this?..which straight lines?...

U r question itself is ambiguous.

from the way I deciphered the answer is 34000
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by scoobydooby » Wed Aug 12, 2009 7:03 am
is it (70C2-20C2+1)C3 ? (since question asks how many triangles out of straight lines)
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by ben wade » Wed Aug 12, 2009 7:08 am
I go with 9500.

for the 20 points in straight line it 20C2, multiplied by 50 for the rest of the points not in the straight line.

so 20C2 * 50 = 9500.

Please post the answer and your logic.
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