15 lines are concurrent at a point P. A line | vedclass

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(D) triangle is formed by selecting any $2$ lines out of the $15$ concurrent lines and the line $L$ acting as the third side.Since the $15$ lines are

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$105$

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(D) triangle is formed by selecting any $2$ lines out of the $15$ concurrent lines and the line $L$ acting as the third side.Since the $15$ lines are concurrent at point $P$,any pair of these lines will intersect at $P$.When these $2$ lines intersect the line $L$ at two distinct points,a triangle is formed with $L$ as one of its sides.The number of ways to choose $2$ lines out of $15$ is given by the combination formula ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$.Thus,the number of triangles is ${}^{15}C_{2} = \frac{15 	imes 14}{2 	imes 1} = 105$.

$15$ lines are concurrent at a point $P$. $A$ line $L$ not passing through $P$ intersects all the $15$ lines and forms triangles with them. Then the number of triangles having $L$ as one of its sides is

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