Finding the maximum number of parallelograms formed if a set of $x$ parallel lines and a set of $y$ parallel lines meet in $12$ points

There is a set of parallel lines with $$x$$ lines in it and another set of parallel lines with $$y$$ lines in it. The lines intersect at $$12$$ points. If $$x>y$$, find the maximum number of parallelograms that can be formed.

My solution approach :-
As there are $$12$$ points which we are getting after the intersection of $$x$$ parallel lines with the $$y$$ parallel lines, then it should mean that $$xy=12$$. Please confirm me on this if I am correct or not?

Now I solved this question diagrammatically i.e. taking $$12$$ points formed by intersection of some $$x$$ parallel lines with the $$y$$ parallel lines and then counting all the possible parallelograms in the diagram and I counted a maximum of $$18$$ parallelograms that can be formed in this condition and my answer is correct !

But I was thinking how this question can be solve mathematically, without doing any manual counting for some bigger numbers of intersection points; let's say there are $$50$$ intersection points or $$1000$$ intersection points? Also is there any importance of $$x>y$$? How it would have impacted the answer if the relation would have been $$y>x$$? Is there a way to solve this question mathematically in bigger number scenarios? Please clarify me on this !!!

• From a solution-verification POV, you have not included enough details. The numerical answer is correct (and if that's all you're going for, great), but the proof part leaves a lot to be desired. EG How do you know that you have considered all possible arrangement of 12 points (esp since that wasn't referenced explicitly). Commented Jun 15, 2021 at 21:51

Given: The lines intersect at $$12$$ points

Therefore $$x \cdot y = 12$$. Since $$x > y$$ possibilities are $$x = 6$$ and $$y = 2$$ or $$x = 4$$ and $$y = 3$$.

We need $$4$$ points to form a parallelogram, $$2$$ points forming $$1$$ line and the other $$2$$ points forming the other line. When $$x = 6$$ and $$y = 2$$, the number of parallelograms formed $$= {}_6C_2 \cdot _2C_2 = 15 \cdot 1$$

When $$x = 4$$ and $$y = 3$$, the number of parallelograms formed $$= _4C_2 \cdot _3C_2$$

Compute this and get the answer