On the sides AB, BC, CA of a △ABC, 3, 4, 5 distinct points (excluding vertices A, B, C) are respectively chosen. The number of triangles that can be constructed using these chosen points as vertices are:
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Official solution
Total points: 3+4+5=12. Choose any 3: (312)=220, minus collinear triples on each side:
\[220 - \binom33-\binom43-\binom53 = 220-1-4-10 = 205\]