Competition problem — Combinatorics

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Combinatorics Difficulty 5.0 AIME, harder Find the answer

A dot is marked at each vertex of a triangle $A B C$ . Then, 2,3 , and 7 more dots are marked on the sides $A B, B C$ , and $C A$ , respectively. How many triangles have their vertices at these dots?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Altogether there are $3+2+3+7=15$ dots, and thus $\binom{15}{3}=455$ combinations of 3 dots. Of these combinations, $\binom{2+2}{3}+\binom{2+3}{3}+\binom{2+7}{3}=4+10+84=98$ do not give triangles because they are collinear (the rest do give triangles). Thus $455-98=357$ different triangles can be formed.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.