= Solution
The <Erdős-Ko-Rado theorem> states that for $1\le r\le n/2$, an <intersecting family> $\mathcal F$ of $r$-subsets of $[n]$ satisfies
$$
\boxed{|\mathcal F|\le\binom{n-1}{r-1}}.
$$
The bound is sharp: take all $r$-sets containing one fixed element.
Use the <Katona circle method>. In any cyclic ordering of $[n]$, the <cyclic interval intersection bound> permits at most $r$ members of $\mathcal F$ to appear as length-$r$ intervals. To see the bound directly, rotate a chosen interval so that it ends at $n$. Intervals ending at $r,\ldots,n-r$ miss it. Pair the remaining endpoints except $n$ as $(j,j+n-r)$, $1\le j\le r-1$; the two intervals in each pair are disjoint, so at most one is chosen. Including the fixed interval gives at most $r$.
There are $(n-1)!$ oriented cyclic orders. A fixed $r$-set appears consecutively in $r!(n-r)!$ of them: collapse it to a block, cyclically order that block with the other $n-r$ elements, and order its members internally. <Double counting> the compatible family-member/order pairs gives $|\mathcal F|r!(n-r)!\le r(n-1)!$, proving the theorem.
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