Lines arrangement

General solutions for n lines is given by $\frac { n\left( n-1 \right) \left( n-2 \right) \left( n-3 \right) }{ 8 }$ .

The question states that no lines are parallel & no three lines are concurrent. Then, the total points of intersections of n such lines is given by $\large \binom {n} {2}$ . Let this number be P.

The total lines formed by joining P such points are $\large \binom {P} {2}$ . But that doesn't give us our answer. Why?

Well, because not every line we counted is 'fresh'!

Note that on every line, there lies ( n-1) points of intersections made by the remaining (n-1) lines. On joining such points, we get a 'stale' line ( or an already existing line).

So, to obtain the answer we subtract the number of such lines from P.

On 1 line consisting of (n-1) intersections, $\large \binom {n-1} {2}$ existing lines can be formed. Thus, on n such lines -- $\large \binom {n-1} {2} * n$ } exisitng lines can be formed.

Our answer is then given by computing $\large \binom {P} {2}$ - n $\large \binom {n-1} {2}$ .