Always remember the basic rule which you missed out that in a triangle sum of two sides must be greater then third side then only a triangle can be formed so there will be more than 4 triangles to be subtracted!!

awesomeusername wrote:A bit tricky.

There are 9 points in the restricted plane. There are three points to a triangle.

9C3 = 9!/3!*6! = 7*8*9/6 = 84

There are four 3 point sets that don't create triangles (when x is the same for all points, or y is the same for all points).

So 84-4 = 80

The correct answer is 76.

Thank you guys

Man, thats a gr8 question
which category does the question fit in, 750+ !?

Clue: We need to discard the combinations which yield Area=0
1. Arrange the 9 coordinates in the matrix
2.9C3
3.subtract 3 horizontal and 3 vertical rows and columns
4.subtract the 2 diagonals
5.9C3-8=76

olylo wrote:The correct answer is 76.

Thank you guys

I found 75, can you make me correct if it is wrong?

There are 84 points as everybody said, then i think we have to reduce 9 three points:

1)(0,1), (0,2), (0,3)
2)(1,0), (2,0), (3,0)
3)(0,1), (1,1), (2,1)
4)(1,0), (1,1), (1,2)
5)(0,3), (1,2), (2,1)
6)(0,3), (1,2), (3,0)
7)(0,3), (2,1), (3,0)
8)(1,2), (2,1), (3,0)
9)(2,0), (1,1), (0,2)

So it will be 84-9=75

9C3-6-2=76

3 Rows+3 Columns of the matrix=6
2 diagonals of the matrix=2

Brent@GMATPrepNow wrote:

Negin wrote:

olylo wrote:The correct answer is 76.

Thank you guys

I found 75, can you make me correct if it is wrong?

There are 84 points as everybody said, then i think we have to reduce 9 three points:

1)(0,1), (0,2), (0,3)
2)(1,0), (2,0), (3,0)
3)(0,1), (1,1), (2,1)
4)(1,0), (1,1), (1,2)
5)(0,3), (1,2), (2,1)
6)(0,3), (1,2), (3,0)
7)(0,3), (2,1), (3,0)
8)(1,2), (2,1), (3,0)
9)(2,0), (1,1), (0,2)

So it will be 84-9=75

I think you missed the restrictions on the values of x and y (1≤x≤3 and 1≤y≤3)

Cheers,
Brent

Thanks Brent.
You are right, I got it now.

Nice explanation...

there are 9 points available to be chosen for triangles to form that can be done in 9C3 ways.
But there are total 8 ways in which three points are colinear and do not form a triangle. Hence the no of triangles = 9C3-8 = 84-8 = 76.
Hence (C) is the correct answer choice.

AkshayaChandan wrote:I have selected the option E as 84 is highest count of triangle in option. The logic behind it is for the given limits of value we can have infinite number of real number combinations. for eg keeping Y fixed x co-ordinate can have the value between 1 to 3. It isn't a mandatory condition that we should select whole number. Is it?

I think if you read the question again, you'll find that it says integer co ordinates.

(B)76

Hi

The proper explanation should be provided by the instructor.Otherwise its very confusing!!

Hi Brent,

That was actually tricky question and good one also.
The first past is quite simple of choice 3 points out of 9 i.e 9C3=84.

but comming to the second point where we have to substract the number of point that
are in one line.
Do we have have kind of formula or logic.
because if we have condition like 1≤x≤10 and 1≤y≤10
then how namy number of points we have to substract.


Please help me to solve it.

Thanks
Nawneet