Proof of Fisher's inequality in combinatorial terms

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

• This is shown in the answer to your earlier question mathoverflow.net/questions/266511/…. – Tom De Medts Apr 6 '17 at 10:53
• I don't see how, can you give me a hint? – Dominic van der Zypen Apr 6 '17 at 11:58
• What do you mean? It's literally there: "At first, we denote $|\mathcal{C}|=m$ and do not assume for a moment that $m=n$, but prove that $m \leq n$." – Tom De Medts Apr 6 '17 at 14:39
• You're right -- I just missed it. Thanks for the hint – Dominic van der Zypen Apr 9 '17 at 7:46

• It looks like that paper deals primarily with $\lambda$-linked designs, which is a condition not imposed above. Can you comment on whether the mentioned (and elsewhere proved) Theorem 3 does apply here? Gerhard "Not Clear If It Does" Paseman, 2017.04.06. – Gerhard Paseman Apr 6 '17 at 16:41