Solution to Problem 2

To answer the first question, observe that for any group of 5 scientists, there must be at least one lock they cannot open. Moreover, for any two different groups of 5 scientists, there must be two different locks they cannot open because if both groups cannot open the same lock then there is a group of 6 scientists among these who will not be able to open the cabinet. Thus, at least C(11,5) = 462 locks are needed.

As to the number of keys each scientist must carry, let A be one of the scientists. Whenever A is associated with a group of 5 other scientists, A should have the key to the lock(s) that these 5 scientists were not able to open. Thus A carries at least C(10,5) = 252 keys.

Although we have only shown that these are lower bounds on the numbers of locks and keys, a scheme can be designed using these many locks and each scientist carrying these many keys.