As to how many locks, for any group of 2 scientists there must be at least one lock they cannot open.
Moreover, for any two different groups of 2, the:”inoperable” lock must be different for each group.
Thus, at least C(4, 2) = 6 locks are needed.
As to how many keys, whenever scientist S is associated with a group of 2 other scientists,
scientist S must have the key to at least one lock that the other 2 scientists cannot open.
Moreover, for any two different groups of 3, the “inoperable” key must be different for each group.
Thus, at least C(3, 2) = 3 keys are needed.
Now, we must design a scheme with 6 locks and 3 keys for each scientist that will actually
accomplish the desired security. For each of the 6 groups of 2, label a lock with the two names on it.
This means that a lock has 2 names on it of those scientists who cannot open it. For each group of
2 scientists bring in a 3rd scientist. He must have the key(s) to the lock(s) that the other 2 scientists
do not have. This will meet the security requirements.
Consider the following scheme to accomplish the desired security:
Denote scientists: a, b, c, d
Denote locks: L1, L2, L3, L4, L5, L6
Labels for locks:
L6(a, b): lock 6 cannot be opened by scientists a, b
L5(a, c): lock 5 cannot be opened by scientists a, c
L4(a, d): lock 4 cannot be opened by scientists a, d
L3(b, c): lock 3 cannot be opened by scientists b, c
L2(b, d): lock 2 cannot be opened by scientists b, d
L1(c, d): lock 1 cannot be opened by scientists c, d
Labels for scientists:
a(L1, L2, L3): scientist a has keys for locks 1, 2, 3
b(L1, L4, L5): scientist b has keys for locks 1, 4, 5
c(L2, L4, L6): scientist c has keys for locks 2, 4, 6
d(L3, L5, L6): scientist d has keys for locks 3, 5, 6