# Combinatorics: Number of subsets of $[n]$ so that each two have a common element [closed]

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?

## closed as off-topic by Aqua, Leucippus, Namaste, Moishe Kohan, DemophilusNov 12 '17 at 1:50

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• What set is [n]? It looks like a modulo class. But of what? [n] = { n + j*M} but what is M? And what is k. n >= 2k???? what is k? – fleablood Nov 11 '17 at 16:35
• This is the famous Erdos-Ko-Rado Theorem. There is a very short, nice proof with the probabilistic method. – jlammy Nov 11 '17 at 16:37
• [N] is a set of n elements – Luka Markovic Nov 11 '17 at 16:42
• "A collection, F, of $k \in X$, so that each two in F have a property". Doesn't actually make any sense. You can't collect all of something with a property that will depend upon which of the things you collect. If I try to select all sets with 1 element so that any two have an elment in common, the I can only pick 1 set, but which one depends upon... which one I pick. So I can't select all of them (because then any two will not have an element in common. – fleablood Nov 11 '17 at 16:43
• Not, "all" subsets of $[n]$ but a collection of some subsets of [n] so that that condition is met. By the way, I have never seen the notation [n] to mean a set of n elements before in my life. – fleablood Nov 11 '17 at 16:45

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 $$\frac{k}{n}\geq\mathbb P(S\in\mathcal F)=\frac{|\mathcal F|}{\binom{n}{k}}\implies|\mathcal F|\leq\binom{n-1}{k-1}.$$