I have an addiction of sorts - I can't keep from playing Freecell. (Most "Windows" users have access to this game from their "Start" menu.) There was once a theory that every possible deal is winnable in this game (this has apparently been disproven).
How many essentially different deals are there in Freecell?
Freecell setup: deal a standard 52 card deck out into eight columns: four with seven cards and four with six. Two deals with only column order changed (i.e., that can be made identical by only switching the locations of particular columns) are not considered different in this context.
(In reply to re: Solution
I agree with the basic solution of 52!/(4!^2).
However, complications ensue if we accept Tristan's alternative
interpretation, whereby Red and Black can be switched without change,
so can hearts and diamonds, clubs and spades. One cannot simply
divide by 2^3, because this undercounts the total number of different
Consider, for instance, if all the black suits are in sequence within
suit. Column 1 = A-7 of spades, column 2 = A-7 of clubs, column 5
= 8-K of spades, column 6 = 8-K of clubs. We can already swap
suits by swapping columns, and we have already accounted for all column
swaps. So we do not need to reduce the total number of deals
further to allow for black suit symmetry in this case.
I'm not sure how much 52!/(4!^2*2^3) undercounts by, but I am sure that
accounting for suit symmetries, the total number of essentially
different deals is something more than 52!/(4!^2*2^3).