If there are $10$ coplanar points,out of which $5$ are collinear,how many total triangles can be formed by joining these points?

Suppose $t_n$ is the number of triangles formed using the vertices of a regular polygon of $n$ sides. If $t_{n+1} = t_n + 28$,then $n =$

The number of diagonals of a polygon is $35$. If $A$ and $B$ are two distinct vertices of this polygon,then the number of all those triangles formed by joining three vertices of the polygon having $AB$ as one of its sides 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:

If there are $m$ parallel lines in each of the two sets of parallel lines that form a parallelogram,then the total number of parallelograms formed is:

The straight lines $l_1, l_2, l_3$ are parallel and lie in the same plane. $A$ total number of $m$ points are taken on $l_1$,$n$ points on $l_2$,and $k$ points on $l_3$. The maximum number of triangles formed with vertices at these points is: