Interesting combinatorics question As a child I had a deck of cards which consisted of maybe 50 cards. On each card there were 6 objects. Players had to pick 2 cards from the deck, and the player who first spotted the difference between the 2 cards won.  The interesting thing was, that for evry 2 cards it was true that they only differed in 1 object. 
Can someone explain how to construct such cards?
 A: One trivial way to construct such cards is to use a total of $55$ objects.  Let's label the objects $A,B,C,D,E,1,2,\ldots,50$.  Each card shows a set $\{A,B,C,D,E,k\}$ with $1\le k\le 50$.  But this is almost certainly not what the OP wants!
A: In response to my comments that Barry Cipra's solution is the only way to construct the deck with only one object different, I'm posting a short proof of that fact.
Consider a deck of $50$ cards that has the property that every card differs in only one object from any other card and is not built where every card has $5$ objects the same and differs in only one spot (so not Barry's deck). Pick two cards at random and call the objects on them $A,B,C,D,E,F$ and $A,B,C,D,E,G$. Using only the so far named objects ($A,B,C,D,E,F,G$), it can be shown that there are only $7$ cards possible using these objects. Because the deck contains more than $7$ cards, there is a card that uses a new object, call it $H$. Pick any $5$ of the already used symbols and add $H$ to the card. This new card (say it is $A,B,C,D,E,H$ for easy reference) must differ in $2$ spots from $5$ of the $7$ already constructed cards. This is a contradiction to the assumptions. So either the deck is really built like Barry has suggested or the deck can not force every pair of cards to be different in only one symbol.
