If 110 triangles can be formed by joining 10 | vedclass

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(A) The total number of ways to select $3$ points out of $10$ is given by $^{10}C_3$.If $n$ points are collinear,they cannot form a triangle,so we sub

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$5$

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(A) The total number of ways to select $3$ points out of $10$ is given by $^{10}C_3$.If $n$ points are collinear,they cannot form a triangle,so we subtract the number of ways to choose $3$ points from these $n$ points,which is $^nC_3$.The number of triangles formed is $^{10}C_3 - ^nC_3 = 110$.Calculating $^{10}C_3 = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120$.So,$120 - ^nC_3 = 110$.$^nC_3 = 120 - 110 = 10$.We know that $^5C_3 = \frac{5 	imes 4 	imes 3}{3 	imes 2 	imes 1} = 10$.Therefore,$n = 5$.

If $110$ triangles can be formed by joining $10$ points as vertices,in which $n$ points are collinear,then the value of $n$ is:

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