My 2 cents:
I agree with the 6c3 explanation. It should be 5C3 when the center point is removed.
If we extended this problem to a square ABCD or a rectangle ABCD and denote the center as E(point of intersection of diagonals), then lets ask the question: how many triangles can be formed with A,B,C,D,E.
This gets a little tricky because now we have 3 points on a straight line: A,E,C and B,E,D - these 2 sets of points cannot form a triangle.
So the solution for a the square/rectangle is:
5C3 - 2 = 8 ways.
Think of this in simple terms:
Say we form triangles without the center point: There are 4C3 = 4 ways of doing this.
Now we form triangle including the center point E. I can select the 2nd vertex of the triangle in 4 ways (A/B/C/D). The 3rd vertex can come only from 2 other points. So total = 4*2 = 8. But this has double counted triangles: eg - EAD, EDA. So we divide by 2. # of triangle with center = 4.
So total = 4+4 = 8.
Shoot questions if you have any.
