There are 6 points on the plain. Any 3 points of these 6 don't lie on the same line. How many unique triangles can be drawn using these 6 points as vertices?
Why isn't the answer 60. We can choose 1 vertice and there are 5C2 = 10 ways to 2 choose the remain 2 vertices. 10*6 choices = 60 (10 choices each for each vertex)?
First of all, please put the OA in stealth form in the original post so everybody is aware of it.
Since the stem says that any 3 points are not in the same line, I simply used simple combination:
C6,3 = 20 triangles