How many triangles can be drawn by means of $9$ non-collinear points?

Let $ABC$ be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side $AB$, five points $p_5, p_6, p_7, p_8, p_9$ on the side $BC$, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side $AC$. None of these points is a vertex of the triangle $ABC$. The total number of pentagons that can be formed by taking all the vertices from the points $p_1, p_2, \ldots, p_{13}$ is . . . . . . .

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 greatest possible number of points of intersection of $8$ distinct straight lines and $4$ distinct circles is

The crew of an $8$-oar boat is to be chosen from $12$ men,of whom $3$ can row on the stroke side only. The number of ways in which the crew can be arranged is

If a polygon of $n$ sides has $275$ diagonals,then $n$ is