alice and bob do a card trick

the victim takes a pack of 52 cards and chooses any five whatsoever. he hands these cards to Bob. Bob looks at the cards. one by one, he takes four of them and hands them to Alice in a certain order. Alice examines the cards Bob has given her and the order they were received in. she then tells the victim what the five cards he chose are.

how did they do it?

It's got to be that there's a way of ordering four random cards that discloses the suit and number of another card. But tell me more...
ways of ordering four cards: 4! = 4x3x2x1 = 24
You gave a hint there though.

Since there are 4 cards out of the pack you only need to nominate one of 48 cards. Each card is paired with a number.

A number Y between 1 and 24 is determined (perhaps using reference tables) by the order of the cards.

A number X = either 0 or 1 is determined by
X=0 if all suits different,
X=1 if two cards same suit,
X=0 if two pairs of cards same suit,
X=1 if three cards same suit,
X=0 of four cards same suit.

It's not pretty. But it is an answer.
Doesn't work if you start with 5 cards in same suit.
it is pretty. but it's not an answer.

doesn't work with 3,1,1 either, since you have to leave either 3 the same or 2 the same and the other 2 different.
Remarkably, it can be done with 124 cards! I should acknowledge that this is not my own solution and was in a paper published by Michael Kleber in 2002. (The Best Card Trick, Mathematical Intelligencer 24#1)

Order them, and number each with 0 - 123. (with standard cards this may be done by ordering suits)

Suppose you have picked cards x_i with x_1 < x_2 < ... < x_5

take k = x_1 + x_2 + ... x_5 mod 5 and hide x_k, give the others.

The message now consists of 4 cards, suppose they sum to X mod 5

This means that the hidden card is congruent to k-X if it is x_k.

Note that 120 cards remain in the pack, if we renumber these with 0-119 then for each ignored x_i < x_k the value of x_k mod 5 (in the new count) decreases by one. Thus the hidden card is congruent to -X mod 5 in the new numbering.

There are now 24 possibilities for the hidden card (120 = 24 x 5). However we may arrange the remaining four cards in any way we wish - so assign each ordering uniquely.
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