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?
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How many triangles on the coordinate plane
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tomada
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The question specifies integer coordinates
I'm really old, but I'll never be too old to become more educated.
When you consider 9C3, that means you want to calculate the number of sets of 3 different points among 9 different points.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.?
Then, the combination you're talking about is not included in the 9C3.
Here the error is in the blodface sentence. After selecting two points among the nine available, we don't always have 6 points left to form a triangle. We could have even 7 points to consider, since the first two points chosen are not always on "the same line" ==> for example, points A(1,1) and B(3,2) are not "on the same line" since we have 7 points left to form a triangle with.wydadi wrote:There are 9 points on the limited plane.
Each point can form 8 couple of points with the 8 remaining points.
If we consider the couple (A,B), we have 7 points left to form an (A,B,C) group. This group is a triangle only if C is not on the same line as A and B ==> means we have 6 points left (since there is one point on the line (AB) that is in the limited plane).
Then we have 9*8*6 groupe (A,B,C). Each group correspond to a triangle on the limited plane.
And since order is not important (triangle ABC is the same as triangle BAC), so each triangle (ABC) is repeated 6 times.
The result is (9*8*6)/6=9*8=72.
The answer of Brent is the most accurate.
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kumarkislay
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Agreed. And since there would be 4 such cases when 7 points would be available - [], [], [], [] => total number of possible triangles would be 72 + 4 = 76. This also matches with 9C3 - 3 (horiz lines) - 3 (vert lines)- 2 (diagonals) = 76.Koala wrote:Here the error is in the blodface sentence. After selecting two points among the nine available, we don't always have 6 points left to form a triangle. We could have even 7 points to consider, since the first two points chosen are not always on "the same line" ==> for example, points A(1,1) and B(3,2) are not "on the same line" since we have 7 points left to form a triangle with.wydadi wrote:There are 9 points on the limited plane.
Each point can form 8 couple of points with the 8 remaining points.
If we consider the couple (A,B), we have 7 points left to form an (A,B,C) group. This group is a triangle only if C is not on the same line as A and B ==> means we have 6 points left (since there is one point on the line (AB) that is in the limited plane).
Then we have 9*8*6 groupe (A,B,C). Each group correspond to a triangle on the limited plane.
And since order is not important (triangle ABC is the same as triangle BAC), so each triangle (ABC) is repeated 6 times.
The result is (9*8*6)/6=9*8=72.
The answer of Brent is the most accurate.
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hungpham
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The answer is C: 78
There are 9 points
Total = (9 x 8 x 7)/6 = 84
Not triangles = 6 including [(0,1) (0,2) (0,3)]; [(1,0) (2,0) (3,0)]; [(1,2) (2,1) (3,0)]; [(0,3) (1,2) (2,1)]; [(0,3) (1,2) (3,0)]; [(0,3) (2,1) (3,0)]
Result = 84 - 6 = 78
Question: If each vertex has to satisfy 1<=x<=3 and 1<=y<=3, there are only three points and there is only one triangle.
There are 9 points
Total = (9 x 8 x 7)/6 = 84
Not triangles = 6 including [(0,1) (0,2) (0,3)]; [(1,0) (2,0) (3,0)]; [(1,2) (2,1) (3,0)]; [(0,3) (1,2) (2,1)]; [(0,3) (1,2) (3,0)]; [(0,3) (2,1) (3,0)]
Result = 84 - 6 = 78
Question: If each vertex has to satisfy 1<=x<=3 and 1<=y<=3, there are only three points and there is only one triangle.
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krishnakumar.ks
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krishnakumar.ks
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I got (B) 76 :
Actually, there are 9 available points/vertices.
Every triangle needs 3 vertices.
so choose 3 out of 9, this is:
9!
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(9-3)!3!
which gives 84.
But we have not to take into account all vertices on the same line
so we have 3 lines of x, 3 lines of y + 2 diagnals.
84-8 = 76
Actually, there are 9 available points/vertices.
Every triangle needs 3 vertices.
so choose 3 out of 9, this is:
9!
-----
(9-3)!3!
which gives 84.
But we have not to take into account all vertices on the same line
so we have 3 lines of x, 3 lines of y + 2 diagnals.
84-8 = 76
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quantskillsgmat
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i think ur answer is wrong.
my answer is 54.
becz the reason that u r giving 9c3-4.there are more three point sets that will nt form traingle.3 verical lines 3 horizontal lines and 2 dignols.
my approach is for any set of two points that lie on same line six traingle are possible.so for 9 points 54 traingle are possible.
my answer is 54.
becz the reason that u r giving 9c3-4.there are more three point sets that will nt form traingle.3 verical lines 3 horizontal lines and 2 dignols.
my approach is for any set of two points that lie on same line six traingle are possible.so for 9 points 54 traingle are possible.
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quantskillsgmat
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now i got ryt approach
answer is 76, this is how
9c3-8=76 there are 8 set of collinear point which will nt form traingles.u can consider a square 4 sides and two digonals.they cant form traingle becz they r collinear.
answer is 76, this is how
9c3-8=76 there are 8 set of collinear point which will nt form traingles.u can consider a square 4 sides and two digonals.they cant form traingle becz they r collinear.
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telberrak
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we can choose 9C3 = 84 combinations of 3 points (A,B,C) from the 9 points.
the number of combination of (A,B,C) that are on the same line are 8 (3 horizontals, 3 verticals, 2 diagonals)
the answer is 84 - 8 = 76 triangles
the number of combination of (A,B,C) that are on the same line are 8 (3 horizontals, 3 verticals, 2 diagonals)
the answer is 84 - 8 = 76 triangles
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gmatscrewed
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i do not get it at all! How do you know that there are 9 points how do u count them? can somebody help me i feel very dumb right now.
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pemdas
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the points can be easily plotted on x-y abscess, the points must be integer values (if we count them 9 only)
C(9,3)=9!/(3!*6!)=84
less the solid line combination sets (horizontals, verticals, diagonals) 8
76 triangles
C(9,3)=9!/(3!*6!)=84
less the solid line combination sets (horizontals, verticals, diagonals) 8
76 triangles
Brent@GMATPrepNow wrote:Source: Beat The GMAT Practice Questions
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
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