Three parallel straight lines L_1, L_2 and L_ | vedclass

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(C) Total number of points $n = 5 + 7 + 9 = 21$. \ The total number of ways to select $3$ points from $21$ is ${}^{21}C_3 = \frac{21 	imes 20 	imes

Vedclass · Accepted Answer

$1201$

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(C) Total number of points $n = 5 + 7 + 9 = 21$. \ The total number of ways to select $3$ points from $21$ is ${}^{21}C_3 = \frac{21 	imes 20 	imes 19}{3 	imes 2 	imes 1} = 1330$. \ $A$ triangle cannot be formed if the $3$ points are collinear. \ Collinear points are those lying on the same line: \ Triangles lost from $L_1$: ${}^{5}C_3 = 10$. \ Triangles lost from $L_2$: ${}^{7}C_3 = 35$. \ Triangles lost from $L_3$: ${}^{9}C_3 = 84$. \ Total triangles = $1330 - (10 + 35 + 84) = 1330 - 129 = 1201$.

Three parallel straight lines $L_1, L_2$ and $L_3$ lie on the same plane. Consider $5$ points on $L_1, 7$ points on $L_2$ and $9$ points on $L_3$. Then the maximum possible number of triangles formed with vertices at these points is:

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