Kleber's [Best card trick][1] proceeds as follows:  The mark (audience member) freely selects five playing cards from a standard deck of $52$ and passes these five to the magician's assistant.  The assistant studies those cards, returns one *mystery card* to the mark, and places the remaining four *exposed cards*, face up, in a sequence on a table.  The magician then enters, inspects the sequence of exposed cards, and correctly announces the identity of the mystery card.

The trick works because of clever mathematics.  The five selected cards must contain a suit that is represented by two (or more) cards.  The assistant will choose one of these as the mystery card, and another of the same suit to be the first in the series of exposed cards.  Thus the magician learns the suit of the mystery card.  

Playing cards can be placed in a canonical order ($\clubsuit A, 2, \ldots K, \diamondsuit A, 2, \ldots K, $ etc.), and thus the three remaining exposed cards can be placed in $3!$ possible order sequences.  Thus the assistant can signal six candidate cards, counting from the value of the first card (modulo 13).  That alone will not cover all 12 potential card values.  Thus instead, the assistant selected as the first card in the sequence to be the the one whose value is fewer than six steps before the other of the final suit, which is the mystery card;  that way the $3!$ possible steps ensure that the mystery card can be reached from the first card in the exposed sequence.

**Question**:  How many four-card exposed sequences can arise in such tricks?

  [1]: http://www.apprendre-en-ligne.net/crypto/magie/card.pdf