Covering with most possible equal size subsets having pairwise singleton intersections Assume I have a non-empty finite set $S$ with $x=|S|$. I want to divide the set $S$ into subsets $S_1, S_2, .., S_n$ (Edit: Yes, $S = \cup S_i$, and I'm embarrassed that I forgot to include that) such that:  


*

*$ |S_i| = y, \forall 1 \le i \le n$  (The cardinality of each subset is fixed)  

*$|S_i \cap S_j| = 1, \forall 1 \le i,j \le n, i \neq j$ (Each subset has exactly 1 element in common with any other subset)


Edit: The second condition above excludes the possiblility of a pairwise disjoint division of subsets.
Is there a way to determine the maximum value of $n$ (the number of possible subsets), given $x$ and $y$?

Adding to @hardmath's aside. An interesting family of solutions in addition to the finite projective spaces are the dual finite affine planes (2-d vector spaces) over finite fields: $GF(q)^2$ . In those, it holds that
(a) two distinct points share exactly one line.
(b) each line contains exactly $q$ points, 
(c) each point is incident with exactly $q+1$ lines.
(d) There are $q(q+1)$ total lines, and $q^2$ total points.
Not every two lines intersect, but as the accepted answer's aside mentioned, if we take the dual space, (a) implies (1p), and (c) implies (1u).
So to connect to the original question, the set $S$ is the set of all lines. The subsets would be indexed by each point $i$:
$S_i=\{l \in S | I(i,l)\}$ - the set of lines intersecting with point $i$.
$\forall_i|S_i|=q+1$. And $S_i \cap S_{j}=$the single line connecting $i$ and $j$, so $|S_p \cap S_{p'}|=1$.
So this is a family of solutions with: $x=q(q+1),\space y=q+1,\space n=q^2$ for all $q$ which is a prime power. Not 100% sure this is maximal, though it sounds like it should be.
Edit: Just realized that in the same way that the affine plane is obtained from the projective plane by removing one line and all the points incident with it, so we can continue to remove additional lines and all the points incident with them from the affine plane. Each two remaining points would still be connected with exactly one line, but not every line will contain an equal number of points, but this wasn't a requirement in the original question. This is pretty wasteful, since it removes many points for each line, but maybe still finds solutions.
