# Families of ordered set partitions with disjoint blocks

Let $$C_1,\dots, C_m$$ be a family of ordered set partitions of $$[n]$$ with exactly $$k$$ blocks.

Write $$C_i = \{B_{i1}, \dots, B_{ik}\}$$ for $$i=1,\dots, m$$ where $$B_{ij}$$ are the blocks of the ordered set partition $$C_i$$.

Suppose this family also has the property that for each $$j=1,\dots, k$$

$$B_{1j} \cup \cdots \cup B_{mj}$$

is also a partition of $$[n]$$

Can one determine the maximal number of members in such a family $$m$$, or at least a decent upper bound on $$m$$?

Edit:

It might also be worth noting that if we take $$k=n$$, then $$m=n$$ since this would be equivalent to the existence of a latin square. I am in particular interested in the case $$k=2$$.

We have $$mn=\sum_i\sum_j |B_{ij}|=\sum_j\sum_i |B_{ij}|=kn,$$ thus $$m=k$$.
Answer: $$m=k$$.
Put indeed your blocks $$B_{ij}$$ in a $$m\times k$$ array and then "read" this array:
-- row-wise: any element of $$[n]$$ appears then $$m$$ times.
-- column-wise: any element of $$[n]$$ appears then $$k$$ times.
• you actually multiply 1 by $m$ and by $k$ :) Apr 9 '19 at 17:13