How many parallelograms can be formed by a set of $4$ parallel lines intersecting another set of $3$ parallel lines?

Let $T_n$ denote the number of triangles which can be formed using the vertices of a regular polygon of $n$ sides. If $T_{n+1} - T_n = 21$,then $n$ equals

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

Statement-$1$: If a polygon has $45$ diagonals,then the number of sides is $10$. Statement-$2$: Out of $n$ non-collinear points,$2$ points can be selected in $^nC_2$ ways.

Suppose $t_n$ is the number of triangles formed using the vertices of a regular polygon of $n$ sides. If $t_{n+1} = t_n + 28$,then $n =$

If the sides $AB, BC$ and $CA$ of a triangle $ABC$ have $3, 5$ and $6$ interior points respectively,then the total number of triangles that can be constructed using these points as vertices is equal to ....... .