Suppose we have $x_1$ lines on a plane parallel in one direction ,$x_2$ lines parallel in another direction, $\cdots$,$x_{10}$ lines parallel in another direction. If $x_i$=$i$ for $i$ $\in${$1,2,3,4,5,6,7,8,9,10$} then what is the maximum possible number of intersections of these lines?
$My$ $approach$ - I tried to break the problem into sub cases. To get the maximum number of intersections we have to consider the non-degenerate cases .
$i=\color{red}1$ $\rightarrow$ #$(intersections)=\color{red}0,$
$i=\color{blue}2$ $\rightarrow$#$(intersections)=\color{blue}2,$
$i=\color{green}3$ $\rightarrow$#$(intersections)=\color{blue}2+\color{green}{3(1+2)},$ $$\vdots$$
So generally , we can see, for eg- if $x_{i}=4$, then we have $1$ straight line, $\color{blue}2$ parallel lines different in direction from the previous one , $\color{green}3$ parallel lines different in direction from all the previous lines.
Now for any new line $\color{brown}L$ which is not parallel to the previous lines , it will intersect the 1st parallel line in one point, two 2nd parallel lines in $\color{blue}2$ points and third parallel lines in $\color{green}3$ points. So for four points we get $\color{purple}4(1+\color{blue}2+\color{green}3)$. The collective number of intersections can be given by adding up the previous number of intersections . What I get is that the general sum for $x_{i}=n$ we have $$2+3(1+2)+4(1+2+3)+\ldots+n(1+\ldots+(n-1))$$ many intersections . Can anyone verify this? And are there in methods in competitive high school maths which can solve this problem in more elegant and simple way? Thank you.