Version of Fitch Cheney's card trick Fitch Cheney's card trick is well known.  Alice picks five cards from a deck.  Bob takes them, gives one back to Alice and arranges the other four in some order.  Chuck then enters the room, looks at the four cards and names the fifth.  This would work with a deck of 124 cards.
Erich Friedman's Math Magic puzzle for March 2006 asks this question. Alice picks a hand of $n$ cards, and Bob shows $k$ of them in some order to Chuck who deduces the other $n-k$ cards.  How big can the deck be?
Suppose a single suit of $13$ cards is used.  Alice picks six cards, then Bob selects three of those and arranges them in some order.  Can Chuck possibly deduce the other three cards?  
I am asking those numbers because there is exactly enough possibilities.  Bob shows Chuck three cards in some order, which has $3!{13\choose3}=1716$ ways, and Alice has ${13\choose6}=1716$ possible hands.   How can Alice's $1716$ hands be paired $6-1$ with Bob's $286$ triples?
 A: If you don't need the mapping to be "easy to memorize" then yes the $(13,6,3)$ case can be done.
In fact, this answer is an easy generalization of this which I found in a comment in this related MO post.
We recast the problem as follows: for any $6$-card subset $S$, we need to encode it with a $3$-card sequence $f(S) = (c_1, c_2, c_3)$ where $c_1, c_2, c_3 \in S$.  Now consider a bipartite graph with node sets $X, Y$, where $X$ contains the $6$-subsets and $Y$ contains the $3$-sequences.  There is an edge from $S\in X$ to $(c_1, c_2, c_3) \in Y$ iff $c_1, c_2, c_3 \in S$.  What we want is a matching which covers every $S \in X$.  
In fact we can find a perfect matching: 


*

*Every $S \in X$ connects to $6\times 5 \times 4 = 120$ different $y\in Y$ because that's the number of $3$-sequences with elements in $S.$

*Every $y \in Y$ connects to ${10 \choose 3} = 10 \times 9 \times 8 / 6 = 120$ different $S \in X$ because there are ${10 \choose 3}$ ways to pick the other $3$ elements of $S.$

*Therefore, the graph is actually a $120$-regular bipartite graph.  A simple application of Hall's marriage theorem shows that a perfect matching exists.
