Let l_1 and l_2 be two lines intersecting at | vedclass

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(D) The total number of points is $8$ (excluding $P$). Including $P$,we have $9$ points in total.
To form a triangle,we need to select $3$ non-collin

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$60$

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(D) The total number of points is $8$ (excluding $P$). Including $P$,we have $9$ points in total.
To form a triangle,we need to select $3$ non-collinear points.
The total number of ways to select $3$ points from $9$ points is ${^9C_3} = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = 84$.
However,points on the same line cannot form a triangle.
Points on $l_1$ (including $P$) are $4$ $(A_1, B_1, C_1, P)$. The number of ways to choose $3$ collinear points from these is ${^4C_3} = 4$.
Points on $l_2$ (including $P$) are $6$ $(A_2, B_2, C_2, D_2, E_2, P)$. The number of ways to choose $3$ collinear points from these is ${^6C_3} = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20$.
Total triangles = (Total ways to choose $3$ points) - (Collinear sets on $l_1$) - (Collinear sets on $l_2$)
Total triangles = $84 - 4 - 20 = 60$.

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 points coincides with $P$,then the number of triangles formed by these eight points is:

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