Could we calculate the number of triangles if we didn't have the restrain for it to be a right triangle? For example, if they only told us that the line RP is prallel to the x axis? So, again for P we would have 110 possabilities, and for R 9.
But what about Q? Could it be anywhere in those 110 points, except those that are P and R. Would that then be 108 choices for Q?
Yes, you can use the same approach. In this case the choices for P would still be 110, and that for R would be 9(the 9 points other than R on the horizontal line PR), however, for Q the choices would be now 100, why? Because Q cannot be on the horizontal line PR, because then all three points would be on a straight line, and they would not form a triangle. So the total number of triangles in this case would be 110x9x100=99000.
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2 Comments
Reader Comments (2)
Could we calculate the number of triangles if we didn't have the restrain for it to be a right triangle?
For example, if they only told us that the line RP is prallel to the x axis?
So, again for P we would have 110 possabilities, and for R 9.
But what about Q? Could it be anywhere in those 110 points, except those that are P and R. Would that then be 108 choices for Q?
Yes, you can use the same approach. In this case the choices for P would still be 110, and that for R would be 9(the 9 points other than R on the horizontal line PR), however, for Q the choices would be now 100, why? Because Q cannot be on the horizontal line PR, because then all three points would be on a straight line, and they would not form a triangle. So the total number of triangles in this case would be 110x9x100=99000.