Suppose $n$ is a positive integer. Let ${\cal C}$ be a set of subsets of $X:=\{1,\ldots,n\}$ with the following properties:


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


A version of [Fisher's inequality](https://en.wikipedia.org/wiki/Fisher%27s_inequality) states that $|{\cal C}| \leq n$. There are short proofs relying on Linear Algebra. Is there a purely combinatorial proof of this statement?



Proof of Fisher's inequality in combinatorial terms
A combinatorial proof of a more general inequality is [given](https://pdfs.semanticscholar.org/f8ac/82fa89f026a5ced840f97086f4b40d370bbd.pdf) by Douglas Woodall.


One line proof Fisher's inequality is g[iven by Renaud Palisse](https://www.evernote.com/shard/s24/sh/58162883-54c7-413e-88a0-a20c0b55b081/e8e764a47c3e95ee1f588ac001f9561f) 


*Palisse, Renaud*, [**A short proof of Fisher's inequality**](http://dx.doi.org/10.1016/0012-365X(93)90179-W), Discrete Math. 111, No.1-3, 421-422 (1993). [ZBL0788.04003](https://zbmath.org/?q=an:0788.04003).


*Woodall, Douglas R.*, [**A note on Fisher's inequality**](http://dx.doi.org/10.1006/jcta.1996.2729), J. Comb. Theory, Ser. A 77, No.1, 171-176, TA962729 (1997). [ZBL0878.05011](https://zbmath.org/?q=an:0878.05011).


