M03-23

Official Solution:

If one of two parallel lines has 7 points on it and the other has 8 points, how many distinct triangles can be formed using these points as vertices?

A. 168
B. 196
C. 316
D. 364
E. 455


Note that this is not a Geometry question. While it uses basic knowledge of lines and figures, it is actually a probability and combinatorics question. There are 8 questions within GMAT Prep Focus Edition that use similar principles. Here is one example.

Approach #1:

There are two types of triangles that can be formed:

1. Triangles with two vertices on the line with 8 points and the third vertex on the line with 7 points: \(C^2_8*C^1_7=28*7=196\);

2. Triangles with two vertices on the line with 7 points and the third vertex on the line with 8 points: \(C^2_7*C^1_8=21*8=168\);

Total number of triangles: \(196+168=364\).

Approach #2:

Any three distinct points chosen from the total of \(8+7=15\) points will form a triangle, except for the cases where the three points are collinear.

Therefore, the total number of triangles can be calculated as \(C^3_{15}-(C^3_8+C^3_7)=455-(56+35)=364\). Here, \(C^3_8\) and \(C^3_7\) represent the number of different sets of 3 collinear points possible from the line with 8 points and the line with 7 points, respectively.


Answer: D