Number of points of intersections of n lines $={\mathrm{nC}}_2$

Let $\mathrm{N}={\mathrm{n}}_{{\mathrm{C}}_2}=\frac{\mathrm{n}(\mathrm{n}-1)}{2}$

By joining ‘N’ points we get ${\mathrm{N}}_{{\mathrm{C}}_2}$ straight lines. But since each line intersects other (n-1) lines.

$\mathrm{n}\left({(\mathrm{n}-1)}_{{\mathrm{C}}_2}\right)$ New lines are not possible.

$\begin{array}{rcl}\mathrm{Number} \mathrm{of} \mathrm{fresh} \mathrm{lines}& =& {\mathrm{NC}}_2-\mathrm{n}\left({(\mathrm{n}-1)}_{{\mathrm{C}}_2}\right)\\ & =& \frac{\mathrm{N}(\mathrm{N}-1)}{2}-\mathrm{n}\frac{(\mathrm{n}-1)(\mathrm{n}-2)}{2}\\ & =& \frac{{\displaystyle \frac{\mathrm{n}(\mathrm{n}-1)}{2}}\left({\displaystyle \frac{\mathrm{n}(\mathrm{n}-1)}{2}}-1\right)}{2}-\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{2}\\ & =& \frac{\mathrm{n}(\mathrm{n}-1)}{2}\left(\frac{{\mathrm{n}}^2-\mathrm{n}-2-4\mathrm{n}+8}{4}\right)\\ & =& \frac{\mathrm{n}(\mathrm{n}-1) ({\mathrm{n}}^2-5\mathrm{n}+6)}{8}\\ & =& \frac{\mathrm{n}(\mathrm{n}-1) (\mathrm{n}-2) (\mathrm{n}-3)}{8} \end{array}$