Let $p_n$ denote the total number of triangles formed by joining the vertices of an $n$-sided regular polygon. If $p_{n+1} - p_n = 66$, then the sum of all distinct prime divisors of $n$ is:

$A$ polygon has $54$ diagonals. Then,the number 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 $3$ squares are chosen at random from the $64$ squares of a chess board,then the probability that all of them lie along the same diagonal line is

The total number of possible ways in which a person can go from $A(0,0)$ to $B(3,3)$ if he can travel only to the right or upward along the grid lines is:

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?