TITLE: Proof of Fisher's inequality in combinatorial terms
QUESTION [6 upvotes]: Suppose $n$ is a positive integer. Let ${\cal C}$ be a set of subsets of $X:=\{1,\ldots,n\}$ with the following properties:

all members of ${\cal C}$ contain at least $2$ elements, and $X\notin {\cal C}$; and
$A\neq B\in {\cal C}$ implies $|A\cap B| = 1$.

A version of Fisher's inequality states that $|{\cal C}| \leq n$. There are short proofs relying on Linear Algebra. Is there a purely combinatorial proof of this statement?

REPLY [7 votes]: A combinatorial proof of a more general inequality is given by Douglas Woodall.
One line proof Fisher's inequality is given by Renaud Palisse 
Palisse, Renaud, A short proof of Fisher's inequality, Discrete Math. 111, No.1-3, 421-422 (1993). ZBL0788.04003.
Woodall, Douglas R., A note on Fisher's inequality, J. Comb. Theory, Ser. A 77, No.1, 171-176, TA962729 (1997). ZBL0878.05011.