Bank's Board of Executive Officers A Bank's Board of Executive Officers is composed of a Director, a Deputy Director and four Heads of Sector. The Director decides to install a new vault. Have several locks made and distribute the keys so that:
• Each key opens exactly one lock.
• The safe is only opened if all its locks are opened.
• The Director can open the safe by himself.
• The Deputy Director may only open the safe with one of the Heads of Sector.
• Heads of Sector can only open the vault in groups of three.
a) What is the minimum number of locks that must be placed in the safe to make this scheme possible?
b) If so, how many keys should each have?
Attemp:
The problem is that I have no feedback, I would like to know if you agree with my answer:
A) 5 Locks
B) Director (E) 5, Deputy Director (v) 4, Heads of Sector (a, b, c, d) 2
Lock1 ABVE
Lock 2 BCVE
Lock3 CDVE
4 DAVE lock
5 ABCDE lock
1 to 4 are to lock the bosses to 3 to open together, 5 is to lock the vice to open alone. What do you think?
 A: The director has a key to each lock so in designing the scheme he isn't really relevant.
Since the deputy director cannot open the vault alone, there must be a lock to which he does not have the key.
Moreover, because he can open the vault with any other head, each lock to which he does not have a key must be opened by all heads.
This can be accomplished with a single lock that can be opened by all heads, with the added condition that the deputy director has a key to each other lock.
So what we're really concerned about is the condition on the heads.
There are $\binom43 = 4$ groups of three heads, and each of them must be able to open these remaining locks.
On the other hand, there are $\binom42 = 6$ groups of two heads and none of them must be able to open all of these remaining locks.
For a group of two heads to fail to open a lock, the lock must contain only the remaining heads.
So for instance, in order for $(A,B)$ to fail to open a lock, that lock must contain only the heads $(C,D)$.
So we start with locks
$$
\text{Cannot be opened by $(A,B)$} \implies \text{ Lock contains only $(C,D)$}
\\\text{Cannot be opened by $(A,C)$} \implies \text{ Lock contains only $(B,D)$}
\\\text{Cannot be opened by $(A,D)$} \implies \text{ Lock contains only $(B,C)$}
\\\text{Cannot be opened by $(B,C)$} \implies \text{ Lock contains only $(A,D)$}
\\\text{Cannot be opened by $(B,D)$} \implies \text{ Lock contains only $(A,C)$}
\\\text{Cannot be opened by $(C,D)$} \implies \text{ Lock contains only $(A,B)$}
$$
This would give us $6$ locks.
We can't do better: if we add a head to one of the locks, some group of two heads will be able to open the vault.
On the other hand, if we remove a head from one of the locks, some group of three heads will fail to open that lock (and hence the vault).
With the preceding locks, this makes a total of $7$ locks.
The director has $7$ keys, the deputy director has $6$, and each head has $4$, for a total of $29$ keys.
