A triangle PQR is such that 3 points lie on the side PQ, 4 points on QR and 5 points on RP respectively. Triangles are constructed using these points as vertices. What is the number of triangles so formed?

Explanation

Total number of points = 3 + 4 + 5 = 12. Total possible triplets = \binom{12}{3} = 220. We must subtract triplets of points that are collinear (lying on the same side). Collinear sets are \binom{3}{3}=1, \binom{4}{3}=4, and \binom{5}{3}=10. Number of triangles = 220 - (1 + 4 + 10) = 205.

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