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

Two squares are chosen at random on a chessboard. The probability that they have a side in common 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 . . . . . . .

$A$ polygon has $54$ diagonals. The number of sides of this polygon is

There are $n$ distinct points on a circle. If the number of pentagons that can be formed using these points as vertices is equal to the number of triangles that can be formed,then the value of $n$ is:

If the number of diagonals of a regular polygon of $n$ sides is $104$,then $n=$