Question

There are '$n$' straight line in a plane, no two of which are parallel and no three pass through the same point. Their points of intersection are joined. Then the maximum number of fresh lines thus introduced is

• A. $\frac{1}{12}n{(n-1)}^2(n-3)$
• B. $\frac{1}{8}n(n-1)(n-2)(n+1)$
• C. $\frac{1}{8}n(n-1)(n-2)(n-3)$
• D. $\frac{1}{8}n(n+1)(n+2)(n-3)$

Answer 1

The correct option is B $\frac{1}{8}n(n-1)(n-2)(n+1)$

No. of total points of intersection $=n{l}_2$

$=\frac{n(n–1)}{2}$.

No. of maximum lens $=$ no. of ways of selection of $2$ points out of $\frac{n(n–1)}{2}$ points.

$=\frac{\frac{n(n–1)}{2}\times [\frac{n(n–1)}{2}–1]}{2}$

$=\frac{1}{8}\left[n^2\right(n–1{)}^2–2n(n–1)]$

$=\frac{1}{8}n(n–1)[n^2–n–2]$

$=\frac{n(n–1)(n–2)(n+1)}{8}$