Let there be $k$ professions and let there be $b$ blocks. We have $n$ people who belong to these $k$ professions, and each person belongs to exactly one profession. We want to place these $n$ people in the $b$ blocks.
For the $j$-th block, let $n_{ij}$ be the number of people belonging to profession $i$ in $j$-th block. Note, some of the $n_{ij}$ may be $0$.
Let $n_j$ be the number of people in $j$-th block. Then $\sum_{i=1}^k n_{ij}=n_j$. Further, $n=\sum_{j=1}^b n_j$. Also assume $n_j>0$ for all $j$.
We want to place the $n$ people in the $b$ blocks (ordering unimportant) such that for every $j,j'$ there exists $i$ such that $n_{ij}>0,n_{ij'}>0$.
If $k,b$ are given, what can be the minimum possible value of $n$?
I would love it if someone can help me get the solution. I believe the answer is $n>b+k-1$ (though unsure about it) but cannot really prove it.