Question

There are $\mathrm{n}$ straight lines in a plane, no two of which are parallel and no three passes through the same point, Their points of intersection are joined. Then the number of fresh lines thus obtained is?

• A. $\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{8}$
• B. $\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)}{6}$
• C. $\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)}{8}$
• D. None of the above.

Answer 1

The correct option is C

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

Evaluate the total number of lines

Given, there are $\mathrm{n}$ straight lines.

Since, no two lines are parallel and no three line passes through the same point, their points of intersection are joined.

Thus, the required number of fresh lines $=\frac{1}{2}\times {}^{\mathrm{n}}\mathrm{C}_2\times {}^{(\mathrm{n}-2)}\mathrm{C}_2$

$$\begin{gathered} =\frac{1}{2}\times \frac{\mathrm{n}\times (\mathrm{n}-1)\times (\mathrm{n}-2)!}{2!\times (\mathrm{n}-2)!}\times \frac{(\mathrm{n}-2)\times (\mathrm{n}-3)\times (\mathrm{n}-4)!}{2!\times (\mathrm{n}-4)!} \\ =\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)}{8} \end{gathered}$$

Hence, option $\left(\mathrm{C}\right)$, $\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)}{8}$ is the correct answer.