The Fano plane is the finite projective plane of order 2, having the smallest possible number of points and lines, 7 each, with 3 points on every line and 3 lines through every point.
Assume that we want to make Fano planes with the numbers from $1$ to $35$.
How many Fano planes can we make in this way?
Notice that for every two Fano planes like $F_1$ and $F_2$ , $F_1$ and $F_2$ must not share any blocks.
Note : What i tried ...
I said we can make $35 \choose 3$ blocks at all ( Call that number $Z$).
Every time that we want to build a Fano plane, We choose $Z \choose 7$ blocks such that the union of these blocks has exactly $7$ members. But the problem is that I don't know how to control this process.
Thanks in advance.
Let $L$ be the set of $35$ lines of $PG(3,2)$. For each plane $\pi$ contained in $PG(3,2)$, define $P_\pi$ to be the set of $7$ lines contained in $\pi$, and for each point $p \in \pi$, define $L_p = \{\ell \in P_\pi:p \in \ell\}$. Then the incidence structure $(P_\pi,L_\pi = \{L_p:p \in \pi\})$ is a Fano plane (axioms follow from the dual statements for $\pi$). Varying this construction over the $15$ planes in $PG(3,2)$ yields $15$ block-disjoint Fano planes on a set of $35$ points.
This is almost definitely not the maximum, but it is a neat construction.