waddap wrote:
I have a questionr regarding this one. Dont we need to divide 364 by 2 as a triangle ACB is the same as ABC etc.? Or am I missing something
Hope below would help.
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. 455
B. 364
C. 196
D. 168
E. 56
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: B