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

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

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$96$

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(C) Total points = $10$. Points that are collinear = $4$. Points that are non-collinear = $10 - 4 = 6$. To form a triangle with at least one vertex from the $4$ collinear points,we consider the following cases: Case $1$: Select $1$ point from $4$ collinear points and $2$ points from $6$ non-collinear points. Number of ways = $\binom{4}{1} 	imes \binom{6}{2} = 4 	imes 15 = 60$. Case $2$: Select $2$ points from $4$ collinear points and $1$ point from $6$ non-collinear points. Number of ways = $\binom{4}{2} 	imes \binom{6}{1} = 6 	imes 6 = 36$. Note: We cannot select $3$ points from the $4$ collinear points because they are collinear and will not form a triangle. Total number of triangles = $60 + 36 = 96$.

There are $10$ points in a plane,of which no three points are collinear except $4$. Then,the number of distinct triangles that can be formed by joining any three points of these ten 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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