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The solution to this is:

One of the 4 suits must be represented 2 times in your set of 5 cards due to the pigeon hole principle. Consider the 13 cards of that suit A=1,... K=13 to be arranged in a clockwise circle. We can see that the cards are at most 6 away from each other, meaning counting clockwise one of them lies at most 6 vertices past the other. Use the "higher" one as the hidden card and place the "lower" one as the first card face up on the table.

Now, using an established value for all 52 cards in the deck, the remaining 3 cards can be placed in 6 orders, Low-Middle-High=1, LHM=2, MLH=3, MHL=4, HLM=5, and HML=6, and give each one of these a value of +1,..., +6 from the first card.

For example, Jc, 2c, 3h, 4d, 2s are handed to you. Choosing the lower club, (2-Jack(11) Mod 13 is 4, so the Jack is the lower one and the 2 is +4 from the jack. Hide the 2c and place the Jc in the first spot. Then using MHL, our +4 value, we arrange the remaining 3 cards with the 3 first, than the 4, now the 2. This implies that the hidden card is a club, and it is +4 from the Jc in the first spot, so we know it is the 2c.

The 124 card solution is discussed in the January 2001 issue of Emissary or Michael Kleberâ€™s article in The Mathematical Intelligencer, Winter 2002 (which I don't have access to).

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# Generalization of Finch Cheney's 5 Card Trick

The Original Fitch Cheney puzzle goes like this:

A Volunteer from the crowd chooses any five cards at random from a deck, and hands them to you so that nobody else can see them. You glance at them briefly, and hand one card bakc, which the volunteer then places face down on the table to one side. You quickly place the remaining four cards face up on the table, in a row from left to right. Your confederate, who has not been privy to any of the proceedings so far, arrives on the scene, inspects the faces of the four cards, and promptly names the hidden fifth card.

It can be shown that with 5 cards there is a strategy to do the trick on a deck of size up to 124 cards (n!+(n-1)).

My question is this: With the audience choosing n cards out of a deck of size (n!+(n-1)) and one card being hidden, how many unique strategies could the magician and the assistant use?