There are 10 points in a plane,out of which 6 | vedclass

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(B) The total number of ways to select $3$ points out of $10$ is $^{10}C_3 = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120$. Since $6$ points

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$N \leq 100$

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(B) The total number of ways to select $3$ points out of $10$ is $^{10}C_3 = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120$. Since $6$ points are collinear,they do not form a triangle. The number of ways to select $3$ points from these $6$ collinear points is $^{6}C_3 = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20$. Therefore,the number of triangles formed is $N = ^{10}C_3 - ^{6}C_3 = 120 - 20 = 100$. Thus,$N = 100$,which satisfies the condition $N \leq 100$.

There are $10$ points in a plane,out of which $6$ are collinear. If $N$ is the number of triangles formed by joining these points,then . . . .

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