# Finding no. of triangles formed on a plane by 20 points

There are 20 points on a plane, no 3 are collinear except 4. Find no. of lines formed, and total no. of triangles that can be formed.

ATTEMPT

No of lines will be $$\binom{20}{2}-\binom{4}{2}+1 = 185$$

No of Triangles using only given points will be : $$\binom{20}{3}-\binom{4}{3} = 1140 - 4 = 1136$$

But the question asks for total no of triangles. so we would also consider triangles with some vertices that are not one of the given twenty points.

So how would we calculate them?

I attempted like so:

A triangle would form if three lines intersect. But these three lines should not intersect at same point. SO we subtract 3-line combinations that meet at same point from total no of 3-line combinations.

so it should be $$\binom{185}{3}-\left[16\times\binom{19}{3}+4\times\binom{17}{3}\right]$$

$$= 1038220-18224 = \boxed{1019996}$$

But i seriously doubt if this is correct. Such a small number can't produce an astronomical no. of Triangles! Please help. Thank you.

EDIT

As pointed by joriki, the answer would depend on position of points, and nothing may be said until that is known. But will my answer hold if no 2 lines are parallel?

• @joriki No i meant in the book it was written maximum. I asked a different question. I didn't mean to ask maximum. Commented Aug 7, 2015 at 10:28