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 `1/8n(n-1)(n-2)(n-3)`

Explanation

Since no two lines are parallel and no three lines 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` (say). 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 the 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)`
`=(1)/(2)xx(1)/(2)n(n-1)[(1)/(2)n(n-1)-1]-(1)/(2)n(n-1)(n-2)`
`=(1)/(8)n(n-1)(n^(2)-n-2)-(1)/(2)n(n-1)(n-2)`
`=(1)/(8)n(n-1)[n^(2)-n-2-4n+8]`
`=(1)/(8)n(n-1)(n-2)(n-3)`