If we'd do it for example for 4-gons, for quadrilaterals, we could start with all the possible quadrilaterals. We could say that the four vertices are a,b,c and d. And then we'd have 6 lines, I mean,
The 4 edges of the quadrilateral, ab, bc, cd and da. And the 2 diagonals, ac and bd.
So if we'd now ignore, out of all the possible quadrilaterals, those who'd have lines of the same length, we'd be left with quadrilaterals with 6 lines of different length.
Now we could order them according to their length. For example: ab < bc < cd < da < ac < bd
So we'd have 6! = 720 possible combinations.
Though it's not exactly 720 possible combinations. Since for any specific quadrilateral, we could have named vertex "a" any of the four vertices of that quadrilateral. So we'd have to go through all 4 of these possibilities too
And we could have placed vertex "b" clockwise or counter-clockwise from vertex "a". So we'd have to go through all 2 of these possibilities too.
If we'd go through all of that for all possible quadrilaterals, would we find all of the possible 720 combinations?
But not all may be possible. So how many would there be? And which would be impossible?
I'm not sure what the answer would be for quadrilaterals. But if we move up to 5-gons, 6-gons, it seems to get more and more complicated.
Has this been researched?
(Sorry for the many edits)
