There are $6$ points on a circle. Two triangles are drawn such that they have no common vertices. What is the probability that none of the sides of the triangles intersect?

In a plane,there are $37$ straight lines,of which $13$ pass through the point $A$ and $11$ pass through the point $B$. Besides,no three lines pass through one point,no line passes through both points $A$ and $B$,and no two lines are parallel. The number of intersection points the lines have is equal to:

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:

Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently equals

There are $m$ points on a straight line $AB$ and $n$ points on another line $AC$,none of them being the point $A$. Triangles are formed from these points as vertices when $(i)$ $A$ is excluded $(ii)$ $A$ is included. Then the ratio of the number of triangles in the two cases is