There are n straight lines in a plane, no two of which are parallel and no three pass through the same point. Their points of intersection are joined. The number of fresh lines thus formed is

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.

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