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It seems that the solitaire card game 'klondike' presents some head scatching when trying to work out odds for winnable games etc, but I would image a 'simpler' question is 'how many different deal combinations are there in the 'Klondike' card game?

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Assuing that you're referring to the standard version of the game as described at Wikipedia, and that the order of the cards in the deck is part of what distinguishes deals, the answer is $52!$, the number of different permutations of $52$ cards. If you disregard the order of the cards in the deck and are only interested in the different ways of dealing the $28$ cards on the table, the answer is $\displaystyle\frac{52!}{(52-28)!}$: There are $52$ choices for the first card to deal, $51$ for the next one, ..., and $52-27$ for the last one.

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Joriki, thanks for that, its left my Excel spreadsheet smoking! Thanks. – Helio Centric Apr 11 '12 at 12:09

I believe joriki is exactly correct, however based on the rules of the game (esp. red-on-black stacking) and the fact that there are 2 red suits and 2 black suits, then the number of truly unique hands (from a playing perspective) can be no more than 26! In other words, for any given hand, if you swap all of the diamonds for each corresponding heart, in the same place in the deal, you will essentially have the same hand, and therefore the same potential permutations of that hand.

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Even if you don't distinguish between the two red and the two black colours (which IMHO isn't completely right because when you put it on the final stack, they are not equivalent), you still have to cards of each kind, so you don't have just $26!$ combinations (which would be with one card of each kind), but $52!/(2!)^{26}$. – celtschk Aug 23 '12 at 22:05

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