OKOSBANK's safe has several different locks. In the bank, $n$ tellers work, each having keys to some of the locks. (A teller can have multiple keys, and multiple tellers can have keys to the same lock.) We know that any $k$ tellers together can always open the safe, but $k-1$ tellers never can. What is the minimum number of locks on the safe?
No matter how we choose $k-1$ treasurers, there is a lock to which none of them has a key; the other $n-k+1$ treasurers each have a key to this lock. Thus, for every $k-1$ element subset of the $n$ treasurers, there is (at least) one lock that none of them can open, and different $(k-1)$-element sets correspond to different locks in this way. The number of locks is therefore at least as many as the number of $k-1$ element (treasurer) subsets, which is $\binom{n}{k-1}$.

We will show that this estimate is sharp, that is, there exists a distribution of keys such that the conditions of the problem are satisfied with $\binom{n}{k-1}$ locks. Number the locks from 1 to $\binom{n}{k-1}$, and label the $k-1$ element subsets of the $n$ treasurers with these same numbers. Consider the following distribution: a treasurer should receive a key to the $i$-th numbered lock if and only if they do not belong to the subset labeled with the $i$-th number. Then no $k-1$ treasurers can open the safe, since none of them has a key to the lock numbered with their set. However, among $k$ treasurers, there is already an owner for each key, since (for every $i$) the number of those without a key to the $i$-th lock is exactly $k-1$.

Ambrus Gergely (Szeged, Radnóti M. Gymnasium, 12th grade)