The sides AB, BC, CA of a triangle ABC have $3, 4$ and $5$ interior points respectively on them. Find the number of triangles that can be constructed using these points as vertices.

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Solution

First note in all 3 + 4 + 5, that is, 12 points are used for forming the triangles. Since 3 points are needed to form a triangle, the total number of triangles (including the triangle formed by collinear points and as these collinear points will not form . triangles their number has to be subtracted from total) is $^{12} C_{3} = 220$ . But this includes the number of triangles formed by 3 points on AB = $^{3} C_{3}$ = 1, the number of triangles formed by 4 points on BC = $^{4} C_{3}$ = 4, and the number of triangles formed by 5 points on CA = $^{5} C_{3}$ = 10.
Hence the required number of triangles
= 220-(1 + 4+10) =205.

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The sides AB, BC, CA of a triangle ABC have 3,4 and 5 interior points respectively on them. Find the number of triangles that can be constructed using these points as vertices.

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