Let $\mathcal F$ be a collection of all subsets of $[n]$ so that each two subsets have a common element and are same size of $k$ ($n \ge 2k$). Prove that $$|\mathcal F| \le {n - 1 \choose k - 1}$$
I have been able to prove that if all those subsets have a common ellement then $\mathcal F = {n-1 \choose k-1}$. But I am struggling with the rest of the problem. Can someone please help?
Lemma. Consider sets $S_j=\{j,j+1,\dots,j+k-1\}$, with addition mod $n$ and $0\leq j\leq n-1$. Then at most $k$ of the $S_j$ are in $\mathcal F$.
Proof. Take some $S_{\ell}\in\mathcal F$. Any other $S_j$ with $S_j\cap S_{\ell}\neq\emptyset$ can be partitioned into $k-1$ pairs $(S_{\ell-j},S_{\ell+k-j})$ with $S_{\ell-j}\cap S_{\ell+k-j}=\emptyset$. The result now follows. $\square$
Now choose a permutation $\pi$ of $\{0,1,\dots,n-1\}$ and some $0\leq i\leq n-1$, at random and independently. Let $S=\{\pi(i),\pi(i+1),\dots,\pi(i+k-1)\}$, with addition mod $n$. Conditioning on $\pi$, by our lemma, $\mathbb P(S\in\mathcal F)\leq\frac{k}{n}$, so $$\frac{k}{n}\geq\mathbb P(S\in\mathcal F)=\frac{|\mathcal F|}{\binom{n}{k}}\implies|\mathcal F|\leq\binom{n-1}{k-1}.$$