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

let AB be any one of n straight lines and suppose it is intersected by some other straight line CD at P.

then, it is clear that AB contains (n-1) of the points of intersection because it is intersected by the remaining (n-1)(n-1) straight lines in (n-1) different points. so, the aggregate number of points contained in the n straight lines is n(n-1). but inn making up this aggregate, each point has evidently been counted twice. for instance. the point P has been counted once among th epoints situated on AB and again among those
on CD.
Hence, the actual number of points`=(n(n-1))/(2)`
Now, we have to find the number of new lines formed by fjoining these points. the number of new lines passing through P is evidently equal to the number of points lying outisde the lines AB and CD for getting a new line joining P with each of these points only.
since, each of the lines AB and CD contained (n-2) points besides the point P, the number of points situated on AB and CD
`=2(n-2)-1`
`=(2n-3)`
`therefore`The number of points outside AB and CD
`=(n(n-1))/(2)-(2n-3)` ltbr the number of new lines pasing through P and similarly through each other points.
`therefore`The aggregate number of new lines passing through the points
`=(n(n-1))/(2){(n(n-1))/(2)-(2n-3)}`
But in making up this aggregate, every line is counted twice. for instance, if Q is one of the points outside AB and Cd, the line PQ is counted once among the lines passing through P hence, actual number of fresh lines introduced.
`=(1)/(2)[(n(n-1))/(2){(n(n-1))/(2)-(2n-3)}]`
`=(1)/(8)n(n-1)(n-2)(n-3)`.