```**Something around 1,600,000,000,000,000.**

If we imagine a plane an arbitrarily far but non-infinite distance in front of me, I can point each of my fingers such that the ray extended from it will intersect that plane at a single point.  Let any parallel rays be considered - for this purpose - to be hitting the same point.  Based on this, I can point my ten fingers to generate a set of dots in an arbitrary order in both the X and Y axes, with points possibly being equivalent along one or both axes.

This describes the following set of options:

If no points are the same, there are 10! ways of ordering along a single dimension = 3,628,800.

If a pair of points are the same, then there are 10 choose 2 = 45*9! orders.  If a triplet are the same, there are 10 choose 3 = 120 * 8! orders.  We continue this sequence on down for a total of (per wolfram Alpha):

sum_2^10binomial(10, X) (11-X)!+10! = 26,065,011

But then we have the cases where one set is already paired off and another set gets paired.  This is the same problem, but with (10 choose 2) * 8 choose from 2 up to 8 with a factorial on the end, noting that we need to divide the first case in half to avoid duplication.  This gives...

binomial(10, 2) sum_2^8binomial(8, X) (9-X)!-binomial(10, 2) binomial(8, 2)×(7!)/2 = 5,436,405

We do this again for 10 choose 3 * 7 choose from 2 up to 7 etc etc:

binomial(10, 3) sum_2^7binomial(7, X) (8-X)!-binomial(10, 3) binomial(7, 3)×(5!)/2 = 2,184,120

And so on and so forth, keeping in mind that this gets trickier as we have more than two clusters, since we get phrases like Bi(10,2)Bi(8,2)Bi(6,3).  I can't do the math at 4:00 am, so I'll approximate that we are at about 40,000,000 total cases.  But this is just one axis - we can do the same on the other axis, for a total of ~ 40,000,000^2 = 1,600,000,000,000,000 distinct situations.  Index them and order to taste.

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