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 then correctly announces the full 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 card values, counting from the value of the first card (modulo 13).  That alone will not cover all 12 potential card values.  This issue is circumvented by the assistant being careful about *which* card of a pair with the same suit is given back and which is used as the first card in the sequence: use as the first card the one whose value is fewer than seven steps before the other of the same suit (modulo 13), 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