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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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Solution my way | Comment 2 of 6 |

The full card deck can be arranged in 52! ways :nterchanging the order of  the 4 sets of seven in the first group of 28 cards reduces this number  4! times , so does interchanging the order of  the 4 sets of six in the remaining part :===>reduces this number another  4! times , 

therefore the answer is 52!/((4!)^2)

  Posted by Ady TZIDON on 2006-09-06 10:08:23
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