posted by Jesse .
There are 20 points all in one straight line and 15 others in a parallel line. How many possible triangles having these points as vertices can be formed?
I can form a triangle by choosing any two points on the one line, then one of the points from the other line.
so the number of ways
= C(20,2)x15 + C(15,2)x20
= 2850 + 2100