Interview Question for Software Development Engineer at Amazon.com:

have to come up with a numbering scheme for each permutation

After encoding the first 20 cards, reindex the remainder and use 5 bits for every card instead of 6. Repeat that.
(20*6 + 16*5 + 8*4 + 4*3 + 2*2 + 1) = 249 bits
32 bytes is 256 bits.

Evert: your solution is good, but can be improved even a little further by starting the encoding with just 1 bit:
1 bit -> encode 1 card
2 bits -> encode 4 cards
3 bits -> encode 8 cards
4 bits -> encode 16 cards
5 bits -> encode 25 cards

1 * 1 + 2 * 4 + 3 * 8 + 4 * 16 + 5 * 25 = 222 bits

@Tester : i dont think you can start with 5 bit code, first indexing has to be 6 bit.
@evert: yours should work!
although the implementation would be a little slow. But with this problem you can achieve the theoretical minimum log_2{52!} (that is the information content of the message passed) by numbering each permutation. That is order (preferably lexicographically) each permutation. And number then incrementally.

@evert, could you please give more detail on how to convert 32 bytes to 52 cards?
Isn't there any conflict with the encodings?
Thanks

huffman coding, I guess

nope, not huffman, there is no frequency variation. everything appears once.

This problem can be handled by using 4 bits for each face card and 2 bits for the suit.
We will use (4*13) for each face card and (2 *4) for the suits.
Combined this would consume 52 for an entire deck of card.

struct Card {
    unsigned char face:4, suit:2;
};