How many triangles on the coordinate plane

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by commonashish » Thu May 30, 2013 12:10 am
but what abt the combination (1,1)(2,2)& (3,3).

?

why cant answer be - 76 ?

pls clarify

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by commonashish » Thu May 30, 2013 12:13 am
but what abt the combination (1,1)(2,2)& (3,3). they alo create the a straight line combination.

?

why cant answer be - 76 ?

pls clarify

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by commonashish » Thu May 30, 2013 12:19 am
but what abt the combination (1,1)(2,2)& (3,3). they alo create the a straight line combination.

?

why cant answer be - 76 ?

pls clarify

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by commonashish » Thu May 30, 2013 12:48 am
but what abt the combination (1,1)(2,2)& (3,3). they alo create the a straight line combination.

?

why cant answer be 76 ?

as there are 8 points that make straight lines.
pls clarify

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by Brent@GMATPrepNow » Thu May 30, 2013 4:28 am
commonashish wrote:but what abt the combination (1,1)(2,2)& (3,3). they alo create the a straight line combination.

?

why cant answer be 76 ?

as there are 8 points that make straight lines.
pls clarify
The answer is 76.

Here's my full solution:

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by tamiri » Thu Jun 06, 2013 3:26 am
There are 9 dots , (x,y) values from which we need to choose 3 to form a triangle. 3C9=84 we have to reduce the dots that are on the same line 3 on y axis, 3 on x axis, 2 diagonals.
Answer = 84-8=76

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by Imsukhi » Mon Jun 24, 2013 7:31 am
Hey! Can come one explain how come there are 9 points in the plane ?

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by Brent@GMATPrepNow » Mon Jun 24, 2013 7:33 am
Imsukhi wrote:Hey! Can come one explain how come there are 9 points in the plane ?
Because we are limiting the vertices to integer coordinates only.
These coordinates, (x,y), must satisfy 1 ≤ x ≤ 3 and 1 ≤ y ≤ 3

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by Java_85 » Sat Aug 03, 2013 1:11 pm
76 is the correct answer.
9 possible points, we need to choose 3 out of 9 where order is not important so: 9!/(6!*3!)=84
now we need to omit those 3 points which can't make a triangle, i.e. 8 of the 84 possible selections, Therefore the right answer is 84-8=76

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by rajeshsinghgmat » Tue Aug 13, 2013 6:25 pm
9C3-3*2-2*1=76

3*2 for 2 sets of 3 parallel lines.

2*1 for the 2 diagonals.

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by ganeshrkamath » Tue Aug 13, 2013 10:20 pm
Brent@GMATPrepNow wrote:How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 1≤x≤3 and 1≤y≤3?
(A) 72
(B) 76
(C) 78
(D) 80
(E) 84
The coordinate range gives a total of 9 points.
So any three points which are not collinear form a triangle.
So the required number = 9C3 - 8 = 9*8*7/6 - 8 = 76
Choose B

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by stevennu » Tue Oct 08, 2013 7:59 am
How would the slot method look in solving this type of question?

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by Brent@GMATPrepNow » Tue Oct 08, 2013 8:32 am
stevennu wrote:How would the slot method look in solving this type of question?
The slot method (aka applying the Fundamental Counting Principle) doesn't really work here because the outcomes of each stage/slot do not differ. For example, selecting (1,2) for the first vertex, is the same as selecting (1,2) for the second vertex.
For more about this, please see https://www.beatthegmat.com/mba/2013/09/ ... s-part-iii

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by b_joy » Tue Oct 22, 2013 7:20 am
There are 9 points in the 1st quadrant. Hence 9C3 combinations are possible with these points.
9C3=84.
But as lines which come on a straight line cannot from a triangle and there are 8 such lines, these must be subtracted from the total.

a b c
d e f
g h i
where (a,b,c), (a, d,g), (g,h,i), (c,f,i), (a,e,i), (c,e,g), (d,e,f) and (b,e,h) are the 8 combinations where a triangle cannot be formed

Hence the answer must be 84-8 = 76

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by KylieAquino87 » Sun Oct 27, 2013 8:12 am
Hi there,

So I realize this post is super old, but I've just began preparing myself for to take the GMAT in 2014 and Quant is what I most need to focus on. Is there anyone who might be able to explain the rationale behind these calculations to me in a way that a person without extensive Quant background could understand?

Thanks a million in advance!