Let l_1 and l_2 be two lines intersecting at | vedclass

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(D) To form a triangle,we need $3$ non-collinear points. The total points are $8$ (excluding $P$).Case $1$: Triangles including point $P$.To form a tr

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$45$

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(D) To form a triangle,we need $3$ non-collinear points. The total points are $8$ (excluding $P$).Case $1$: Triangles including point $P$.To form a triangle with $P$,we must select one point from $l_1$ and one point from $l_2$.Number of ways = $^3C_1 	imes ^5C_1 = 3 	imes 5 = 15$.Case $2$: Triangles not including point $P$.We can select $2$ points from $l_1$ and $1$ point from $l_2$,or $1$ point from $l_1$ and $2$ points from $l_2$.Number of ways = $(^3C_2 	imes ^5C_1) + (^3C_1 	imes ^5C_2) = (3 	imes 5) + (3 	imes 10) = 15 + 30 = 45$.Wait,the total number of ways to choose $3$ points from $8$ is $^8C_3 = \frac{8 	imes 7 	imes 6}{3 	imes 2 	imes 1} = 56$.We must subtract the cases where the $3$ points are collinear (i.e.,all $3$ points lie on $l_1$ or all $3$ points lie on $l_2$).Number of collinear sets on $l_1 = ^3C_3 = 1$.Number of collinear sets on $l_2 = ^5C_3 = 10$.Total triangles = $56 - (1 + 10) = 56 - 11 = 45$.

Let $l_1$ and $l_2$ be two lines intersecting at $P$. If $A_1, B_1, C_1$ are points on $l_1$,and $A_2, B_2, C_2, D_2, E_2$ are points on $l_2$,and if none of these coincides with $P$,then the number of triangles formed by these eight points is:

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