The following exercise is form P. Cameron, Permutation Groups:
2.22 The Mathieu group $M_{23}$ is a $4$-transitive permutation group on the set $\Omega = \{1,\ldots, 23\}$, and preserves a set $\mathcal B$ of $253$ subsets of $\Omega$, each of size $7$. From this information alone, show that $M_{23}$ acts transitively on $\mathcal B$ with rank $3$.
As the group acts $4$-transitive, among the $4$-subsets of the $7$-subsets in any orbit on $\mathcal B$ appears every possible $4$-subset. As each $7$-subsets can account for at most $\binom{7}{4} = 35$ differnent $4$-subsets, we must have all of the $253$ subsets of $\mathcal B$ in one orbit, for otherwise we do not have enough $7$-subsets to ''fit in'' all $4$-subsets. As $253$ is the least possible value that works for these arguments, every $4$-subsets determines uniquely its $7$-subset in which it is contained, this gives $|A \cap B| \ge 4$ implies $A = B$ for $A,B \in \mathcal B$. The rank for this action on $\mathcal B$ is the number of orbits on $\mathcal B\times \mathcal B$, for $(A,B) \in \mathcal B\times\mathcal B$ the cases $|A\cap B| \in \{0,1,2,3\}$ are possible by the above, and they all give distinct orbits on the cartesian product. Hence the rank is at most $4$, but I still need to argue that one orbit could not arise (and the other three are realised). But here I am stuck. So any ideas how to finish the argument to show that the rank is three?