There are 10 points in a plane of which no th | vedclass

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(B) Total points = $10$. Points that are collinear = $4$. Points that are non-collinear = $6$. To form a triangle with at least one vertex from the $4

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$96$

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(B) Total points = $10$. Points that are collinear = $4$. Points that are non-collinear = $6$. To form a triangle with at least one vertex from the $4$ collinear points,we consider two cases: Case $1$: One vertex from the $4$ collinear points and two vertices from the $6$ non-collinear points. Number of triangles = $^4C_1 	imes ^6C_2 = 4 	imes 15 = 60$. Case $2$: Two vertices from the $4$ collinear points and one vertex from the $6$ non-collinear points. Number of triangles = $^4C_2 	imes ^6C_1 = 6 	imes 6 = 36$. Total number of triangles = $60 + 36 = 96$.

There are $10$ points in a plane of which no three points are collinear except $4$ points. The number of distinct triangles that can be formed by joining these points such that at least one of the vertices of every triangle formed is from the given $4$ collinear points is:

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