Question

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. Then the number of fresh lines thus obtained is

• A. $\frac{n(n-1)(n-2)(n-3)}{8}$
• B. $\frac{n(n-1)(n-2)(n-3)}{6}$
• C. $\frac{(n-2)(n-3)(n-4)}{8}$
• D. $\frac{n(n-2)(n-3)}{8}$

Answer 1

The correct option is A $\frac{n(n-1)(n-2)(n-3)}{8}$

No. of intersection points = $\frac{{}^n{C}_2}{2}$ (as we have counted each line twice by paying heed to order. excluding the order we get 1/2 the no. of points.

No. of ways of selecting another intersection point not lying on the respective 2 lines can be found by counting no. of ways 2 lines can be selected together from the remaining set of (n-2) lines since if the point lies on either of the lines, joining the 2 will not give a fresh line.

Hence

No. of fresh lines = $\frac{1}{2}{\times }^n{C}_2{\times }^{n-2}{C}_2$

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