Ten points lie in a plane such that no three of them are collinear. The number of lines passing through exactly two of these points and dividing the plane into two regions,each containing four of the remaining points,is

The number of triangles that can be formed by choosing the vertices from a set of $12$ points,$7$ of which lie on the same straight line,is

The number of straight lines that can be formed by joining $20$ points,no three of which are in the same straight line,except $4$ of them which are in the same line,is:

The number of triangles $(T_n)$ that can be formed using the vertices of a polygon with $n$ sides is given by $T_n = \binom{n}{3}$. If $T_{n+1} - T_n = 21$,then what is the value of $n$?

Out of $n$ points in a plane,$p$ points are collinear. (No three of the remaining points are collinear). The number of lines that can be drawn passing through these points is:

Consider a rectangle $ABCD$ having $5, 6, 7, 9$ points in the interior of the line segments $AB, BC, CD, DA$ respectively. Let $\alpha$ be the number of triangles having these points from different sides as vertices and $\beta$ be the number of quadrilaterals having these points from different sides as vertices. Then $(\beta-\alpha)$ is equal to: