Problem: Fifteen distinct points are designated on $\triangle A B C$ : the 3 vertices $A, B$, and $C ; 3$ other points on side $\overline{A B}, 4$ other points on side $\overline{B C}$; and 5 other points on $\overline{C A}$. Find the number of triangles whose with positive area whose vertices are among these 15 points. Solution: Note that there are $\binom{15}{3}$ ways to choose any three points from among the $15$. Now, these points are *not* a triangle if they are collinear, meaning they fall on the same line. Hence, the "bad" choices are $\binom{5}{3}+\binom{6}{3}+\binom{7}{3}$ because there are five points that lie on $AB$, six on $BC$ and seven on $CA$. Using a complementary counting approach, our final answer is hence $\binom{15}{3} - \binom{5}{3} - \binom{6}{3} - \binom{7}{3} = 390$.