Question

Six lines are drawn, no two of which are parallel. If no more than two of the lines pass through any one point, what is the number of triangles formed?

Answer 1

20

Explanation:

combinations

C(n,r)= n! / (r! * (n−r)!)

n = 6 : total number of lines

r = 3 : because its 3 lines to make a triangle

6 lines

6C3 = 20 triangles

quora glenn clemens

mathplanet

Answer 2

Answer: I believe 20.

Take one pair of lines. They will intersect at a point (no lines parallel). The other four lines will intersect at different points (no more than two lines through one point) giving four triangles. There are 6C2 = 15 different pairs of lines so 15*4 = 60 triangles. However, each triangle will come from three different points so we need 60/3 = 20 distinct triangles.

EDIT: Now that I’m more awake, it occurs to me there is a much easier answer. Since no two lines are parallel and no three lines are coincident, every combination of three lines must form a triangle. There are 6C3 = 20 triangles.

Explanation: