There are $n$ straight lines in a plane, in which no two are parallel and no three pass through the same point. Their points of intersection are joined. Show that the number of fresh lines thus introduced is $\frac{1}{8} n (n - 1) (n - 2) (n - 3)$ .

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Solution

Since no two lines are parallel and no three this pass through the same point, their points of intersection, i.e., number of ways of selecting two lines from $n$ lines is $^{n} C_{2} = N (s a y)$ . It should also be noted that on each line there will be $(n - 1)$ points of intersection made by the remaining $(n - 1)$ lines.
Now we have to find number of new lines formed by these points of intersections. Clearly, a straight line is formed by joining two points so the problem is equivalent to select two points from $N$ points. But each old line repeats itself $^{(n - 1)} C_{2}$ times [selection of two points from $(n - 1)$ points on this line]. Hence, the required number of new lines is
$^{N} C_{2} - n^{n - 1)} C_{2} = \frac{1}{2} \times \frac{1}{2} n (n - 1) [\frac{1}{2} n (n - 1) - 1] - \frac{1}{2} n (n - 1) (n - 2)$
$= \frac{1}{8} n (n - 1) (n^{2} - n - 2) - \frac{1}{2} n (n - 1) (n - 2)$
$= \frac{1}{8} n (n - 1) [n^{2} - n - 2 - 4 n + 8]$
$= \frac{1}{8} n (n - 1) (n - 2) (n - 3)$ .

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