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How many triangles on the coordinate plane

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by kvlpnd » Sun May 10, 2015 11:28 pm
hitmis wrote:Why should we not have to also subtract combinations of (x1,y1) = (x2,y2)=(x3,y3) , 9 such combinations, which would also not create a triangle.?
We dont have to. Because these selections are already removed when calculating 9C3.
Evaluate 9C3
9C3=9!*8!*7!/(3!*2!*1!)

Here, 8and7 shows reduced choices. Hence, repeated choices are already removed.
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by chanstee » Fri Jul 10, 2015 5:23 pm
Good question
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by chanstee » Fri Jul 10, 2015 5:23 pm
Good question
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by chanstee » Fri Jul 10, 2015 5:23 pm
Good question
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by chanstee » Fri Jul 10, 2015 5:23 pm
Good question.
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by axay » Sat Sep 12, 2015 9:52 am
9C3-3-3-2 = 76
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by Kirchoff440 » Sun Apr 17, 2016 2:17 am
Answer is 84 Triangles
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by Brent@GMATPrepNow » Sun Apr 17, 2016 6:28 am
Kirchoff440 wrote:Answer is 84 Triangles
That isn't the correct answer. Keep trying.

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by Goddin » Fri May 27, 2016 12:34 am
IMO: 76 is correct.

Because 84-6(3 points in a straight line)=76.
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by Matt@VeritasPrep » Fri May 27, 2016 2:16 pm
Goddin wrote:IMO: 76 is correct.

Because 84-6(3 points in a straight line)=76.
76 is right ... but we might want to check that arithmetic :)
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TTT

by sumistudy747 » Tue Jan 02, 2018 1:38 am
The answer would be 76. The total number of combinations would be 9c3= 84, so the answer needs to be something less than this. If you draw the figure and mark all possible points on the x and y axis(it would be a rectangle), you will clearly see that there are 8 lines that would be formed(6 horizontal and vertical and 2 diagonal). If we subtract these 8. We get 84-8=76.
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TTT

by DeepaVarma » Fri Jan 12, 2018 1:02 am
Total 3 points' combinations = 9C3

nCr = n! / ((n-r)! . r!)

9C3 = 9! / 6!.3! = 9.8.7 / 3.2 = 84

Out of 84 combination , 8 will be straight lines

84 - 8 = 76
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