two cases possible
1. selecting two points from abcde and joining the two points to F
5c2*1 = 10
2.selecting three points from 5c3 = 10

adding 10 +10 = 20

The answer is definitely C. Here we simply need to calculate how many ways we can use to built a triangle from 6 points or how we can choose randomly 3 points from 6 given ones. The number of points is calculated in such a way:
6!/((6-3)!*3!) = 6!/(3!*3!)=4*5*6/3! = 4*5 = 20

There is a formula from the combinatorial theory that lets you calculate the number of ways how to choose m elements from n - n!/((n-m)!*m!)

Hope this explanation helps

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.

Why do you think 5C3 is incorrect?

shrek1089,

Are you referring to the q. corresponding to a square/rectangle?

If so, I have provided the solution in my previous post. In short, 5C3 takes into account 2 cases where three points lie on a straight line (along the diagonals) and cannot form a triangle.

So correct solution is to subtract 2 from 5C3.