A club provides 30 snacks to 18 members, and each member orders 3 different snacks. It is known that every snack is ordered by at least one member, and that any two members order at most one same snack. Is it possible to find 12 snacks, such that the snacks ordered by any member is not completely in these 12 snacks?
this should have beeen solved long ago.
use a probabilistic argument; consider a random set of $S\subset X$(the set of snacks) of size $12$ and let$1_A(S)=1$ if $A\subset S$, and $0$ otherwise.then
$N=\sum_{A\in\mathcal{A}} 1_A$ counts the number of subsets of $\mathcal{A}$ contained in $S$, so that $E(N)=\sum_{A\in\mathcal{A}} E(1_A)$. now $E(1_A)=P(A\subset S)=\frac{\binom{27}{9}}{\binom{30}{12}}$ and so $E(N)=18\cdot \frac{10\cdot 11\cdot 12}{30\cdot 29\cdot 28}=\frac{198}{203}$ and so there always exists a set of size $12$ with none of the elements of $\mathcal{A}$ contained in it.
We take a snack from each of the 18 ordering lists of the members. There must exist 12 snacks not containing any of the above 18 (possibly repeated) snacks, and those 12 snacks are what we are looking for.