Too many lines there!

Total numbers of point formed by intersection of ' $n$ ' lines $= {n \choose 2}$

A point, say $p$ , formed by intersection of two lines will be collinear with all the points that lie on these two lines. Each of these two lines will have ' $n-2$ ' points other than point $p$ .

$\Rightarrow$ Number of points that are collinear with $p = 2n-4$

$\Rightarrow$ Number of points that are not collinear with $p$ $= {n \choose 2} - (2n-4) -1$

$= \frac{n \times (n-1)}{2} - 2n + 4 -1$

$= \frac{n^2 - 5n + 6}{2}$

$= \frac{(n-2)(n-3)}{2}$

Hence every point makes $\frac{(n-2)(n-3)}{2}$ new lines.

For all ${n \choose 2}$ points, total number of new lines formed $=$

$\frac{1}{2} \times$ Total number of points $\times$ New lines formed by every point $=$

$= \frac{1}{2} \times {n \choose 2} \times \frac{(n-2)(n-3)}{2}$

$= \frac{1}{2} \times \frac{n \times (n-1)}{2} \times \frac{(n-2)(n-3)}{2}$

$= \frac{(n)(n-1)(n-2)(n-3)}{8}$

$= 3 {n \choose 4}$

Each line will be defined by a set of 4 points, and there are (n choose 4) such sets. These 4 points need to be paired off with each other, and there are 3 way of doing that.