You can
put this solution on YOUR website!I think the answer is D - none of the above. I get a total of 105 triangles
I'm assuming coliner means they are on the same line.
5 points on the same line (assume they form one side of the triangle). So how many distinct 2 points (as two points constitue one side) can you select from a set of 5 points. Note side AB = Side BA so this is a combination problem.
5! div 2!*3! = 10 sides -------------- A
From the Remaining 7 points, how many triangles can you make. This is same as asking you to select 3 distinct points from 7. Again a combination problem as Triangle ABC = Triangle BAC
7! div 3!*4! = 35 -------------------- B
NOT DONE YET !!!
Each side in A can combile with each of the 7 points to make a triangle. Thus, 10 sides can make 70 triangles ----- C
Answer = C+B = 105