Question

There are $n$ straight lines in a plane, no two of which are parallel and no three of which pass through the same point. If their points of intersection are joined, so that number of fresh lines thus introduced will be.

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

Answer 1

Answer: (A)

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

There are ${}^n{C}_2$ ways of choosing one point of intersection (pick two distinct lines from the set of $n$ lines);

There are ${}^{n-2}{C}_2$ ways of picking the second point of intersection to determine the new line (pick two distinct lines among the remaining $n-2$ lines).

The two points uniquely determine a new line.

But you've counted each line twice, since the order in which you select the two points does not matter.

So the total number should be

$\frac{1}{2}\times {}^n{C}_2\times {}^{n-2}{C}_2$

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