Explanation

(c)
Since, no two lines are parallel and no three are concurrent, therefore $n$ straight lines intersect at $\ ^{n}C_{2} = N$ (say) points. Since, two points are required to determine a straight line, therefore the total number of lines obtained by joining $N$ points is $\ ^{N}C_{2}$. But in this each old line has been counted $\ ^{n - 1}C_{2}$times. Since, on each old line there will be $n - 1$ lines.
Hence, the required number of fresh lines.
$= \ ^{N}C_{2} - n\ ^{n - 1}C_{2}$
$= \frac{N(N - 1)}{2} - \frac{n(n - 1)(n - 2)}{2}$
$= \frac{\ ^{n}C_{2}(\ ^{n}C_{2} - 1)}{2} - \frac{n(n - 1)(n - 2)}{2}\ \ (\because N = {\ ^{n}C}_{2})$
$= \frac{\frac{n(n - 1)}{2}\left( \frac{n(n - 1)}{2} - 1 \right)}{2} - \frac{n(n - 1)(n - 2)}{2}$
$= \frac{n(n - 1)}{8}\lbrack\left( n^{2} - n - 2 \right) - 4(n - 2)\rbrack$
$= \frac{n(n - 1)}{8}\lbrack n^{2} - 5n + 6\rbrack$
$= \frac{n(n - 1)(n - 2)(n - 3)}{8}$