implicity

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.