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Freecell (Posted on 2006-09-06) Difficulty: 2 of 5
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.

No Solution Yet Submitted by Cory Taylor    
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re(3): Solution complications | Comment 6 of 7 |
(In reply to re(2): Solution complications by Steve Herman)

Yeah, you're right.  I was undercounting it.

I'm pretty sure there's a way to do this without counting through every solution, but I don't remember how because my combinatorics are pretty rusty.  Even if I did remember, I believe it would come out extremely complicated (imagine having to separate the problem into 4!^2 cases).

  Posted by Tristan on 2006-09-11 15:17:00

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