The sides AB, BC, CA of a triangle ABC have r | vedclass

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(A) Total number of points = $3 + 4 + 5 = 12$.The total number of ways to select $3$ points out of $12$ is $^{12}C_3 = \frac{12 	imes 11 	imes 10}{3

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$205$

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(A) Total number of points = $3 + 4 + 5 = 12$.The total number of ways to select $3$ points out of $12$ is $^{12}C_3 = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = 220$.However,points lying on the same side are collinear and cannot form a triangle. We must subtract the cases where $3$ points are selected from the same side:$1$. Points on side $AB$: $^{3}C_3 = 1$ way.$2$. Points on side $BC$: $^{4}C_3 = 4$ ways.$3$. Points on side $CA$: $^{5}C_3 = 10$ ways.Total collinear sets = $1 + 4 + 10 = 15$.Number of triangles = $220 - 15 = 205$.

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

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