*Also asked on MSE: What is the best way to partition the $4$-subsets of $\{1,2,3,\dots,n\}$?.*

Consider the set $X = \{1,2,3,\dots,n\}$. Define the collection of all $4$-subsets of $X$ by $$\mathcal A=\{Y\subset X: Y\text{ contains exactly $4$ elements}.\}$$

I want to partition $\mathcal A$ into groups $A_1,A_2,\dots, A_m\subset \mathcal A$ (each of them is a collection of $4$-subsets of $X$) such that $\bigcup_{i=1}^m A_i=\mathcal A$ and such that the intersection of any two distinct $4$-subsets in each $A_k$ has cardinality at most $1$, i.e. such that for all $i\in\{1,\dots,m\}$ and $Y_1, Y_2\in A_i$, we have $$Y_1\neq Y_2 \implies \lvert Y_1\cap Y_2\rvert \le 1.$$

**My question:** What can be said about the smallest $m$ (depending on $n$) such that such a partition exists?

**My thoughts:** I was expecting that each $A_i$ can contain "roughly" $\frac n4$ elements, so we would have $$m(n)=\Theta\left(\frac{\binom n4}{\frac n4}\right)=\Theta(n^3).$$ In particular, we would have $m(n)\le c n^3$ for some constant $c\in\mathbb R$.

However, I am neither sure if this is correct, nor how to formalize this.