Two lines $l_1$ and $l_2$ intersect at a point $P$. 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 none of these points coincide with $P$,how many triangles can be formed using these $8$ points?

Let $P_{1}, P_{2}, \ldots, P_{15}$ be $15$ points on a circle. The number of distinct triangles formed by points $P_{i}, P_{j}, P_{k}$ such that $i+j+k \neq 15$ is:

If in a regular polygon the number of diagonals is $54$,then the number of sides of this polygon is

Line $L_1$ with slope $2$ and line $L_2$ with slope $\frac{1}{2}$ intersect at the origin $O$. In the first quadrant,$P_1, P_2, \ldots, P_{12}$ are $12$ points on line $L_1$ and $Q_1, Q_2, \ldots, Q_9$ are $9$ points on line $L_2$. The total number of triangles that can be formed having vertices at three of the $22$ points $(O, P_1, P_2, \ldots, P_{12}, Q_1, Q_2, \ldots, Q_9)$ is:

Let $T_n$ denote the number of triangles which can be formed using the vertices of a regular polygon of $n$ sides. If $T_{n+1} - T_n = 21$,then $n$ equals

There are $n$ straight lines in a plane,no two of which are parallel and no three pass through the same point. Their points of intersection are joined. Then the number of fresh lines thus obtained is