Finding set of subsets satisfying certain properties I came across an interesting puzzle question that can be reduced to the following mathematical question.
Consider a set of seven elements $S=\{a, b, c, d, e, f, g\}$.
We want to construct a set of subsets of $S$ given a few conditions.
(1) For any two different elements in $S$, there is exactly one subset containing them.
(2) Any two distinct subsets $S_1, S_2$ of $S$ satisfy $S_1\cap S_2=\{x\}$ for some $x\in S$. 
(3) The cardinality of the set of subsets is strictly less than $7$.
Also, we assume that $S$ contains at least four elements such that no three of them belong to the same subset.
The question is, does such a set of subsets exist?
I am having trouble getting an answer. Any help is appreciated.
 A: No, it's not possible (as Shuri2060 points out, having the whole of $S$ as a subset breaks the extra fourth rule). 
First, suppose there is a subset of size $4$, say $\{a,b,c,d\}$. We must have other subsets which cover the eight pairs $a,f$, ..., $d,f$ and $a,g$, ..., $d,g$. Now no subset can cover more than two of these without having a pair in common with the first subset, and any subset which does cover two of these eight pairs must include $f$ and $g$, so at most one subset covers two of them. Thus there would have to be at least seven extra subsets to cover all these pairs. So we can't do it if any subset has size $4$. Exactly the same argument would apply if our starting subset was $\{a,b,c,d,e\}$ of size $5$. If we have a subset of size $6$ then there are $6$ other pairs to cover, and each other subset can cover at most one without covering a pair that's already done, so we need at least seven subsets in total.
On the other hand, if every subset has size at most $3$ then each subset covers at most three pairs, and there are $21$ pairs to cover, so at least $7$ subsets are required. (It is possible with exactly $7$ subsets of size $3$: take $abf$, $ace$, $adg$, $bcd$, $beg$, $cfg$, $def$. This construction is the Fano plane.)
