There are 20 straight lines in a plane such that no two of them are parallel and no three of them are…
There are 20 straight lines in a plane such that no two of them are parallel and no three of them are concurrent. If their points of intersection are joined, then the number of new line segments formed are
3420
14535
2907
17955
Solution
We know that, if there are $n$ straight lines in a plane, no two of which are parallel and no three passes through the same point and their point of intersection are jointed then number of fresh lines thus introduced is
$
=\frac{1}{8} n(n-1)(n-2)(n-3)
$
$
\begin{aligned}
& =\frac{1}{8} \times 20 \times(20-1) \times(20-2) \times(20-3) \\
& =\frac{1}{8} \times 20 \times 19 \times 18 \times 17 \\
& =\frac{116280}{8}=14535
\end{aligned}
$