$A$ regular polygon of $n$ sides has $170$ diagonals,then $n$ is equal to

$110$ triangles can be formed by joining $10$ points as vertices,in which $n$ points are collinear. Then the value of $n$ is:

The number of diagonals in a polygon is $20$. The number of sides of the polygon is:

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 . . . . . . .

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

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?