TITLE: Lock combinatorics
QUESTION [7 upvotes]: 'Six generals propose locking a safe containing top secret information with a number of different locks. Each general will be given keys to certain of these locks. How many locks are required and how many keys must each general have so that, unless at least four generals are present, the safe cannot be opened?'
I suspect this question is more complicated than my interpretation of the language. To me the answer seems to be that if each general is given n keys, there must be 3n+1 locks. This is because in the 'worst-case' scenario where each has n distinct keys, 3 generals can open 3n locks.
Perhaps this is a gross simplification, but if anyone cares to comment that would be a great help!

REPLY [3 votes]: Hint: Label the generals "A, B, C, D, E, F".
Label the locks with a subset of $\{A, B, C, D, E, F\} $.
Each general (say A) has a key to lock $L$, if the lock has a label of $A$.
Find a simple configuration which does what we want.
Further Hint: Consider all subsets of 3 elements. Clearly, if we have 3 generals, they cannot open all the locks. If we have 4 generals, they can open all the locks. This shows that ${6 \choose 3} = 15$ locks are sufficient. Each general has ${5 \choose 2} = 10 $ locks.
Further further hint: How do we show that 15 locks is sufficient? Claim: Given any 3 generals, there is a unique lock that they cannot open. Hence done.