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Claim. For $k=2$ the conjecture of Problem 1020 asks for the largest family of $r$-subsets of an $n$-set with no two disjoint members, that is, the largest intersecting family. Theorem 1 of Erdős, Ko and Rado, in the problem's terms, states that for $n\ge2r$ an intersecting family of $r$-subsets of an $n$-set has at most $\binom{n-1}{r-1}$ members, with equality for the family of all $r$-sets through a fixed point; the paper's Remark records that the bound is attained. Since $\binom{n-1}{r-1}=\binom nr-\binom{n-1}{r}$ and $\binom{n-1}{r-1}\ge\binom{2r-1}{r}$ for $n\ge2r$, with equality at $n=2r$, this is

$$f(n;r,2)=\max\left(\binom{2r-1}{r},\binom nr-\binom{n-1}{r}\right) \qquad(n\ge2r),$$

the conjectured value at $k=2$ over the whole range $n\ge2r$ of the corrected Statement. For $n\le2r-1$ any two $r$-sets meet, so $f(n;r,2)=\binom nr$; these values lie outside the corrected Statement. The paper is P. Erdős, Chao Ko and R. Rado, Intersection theorems for systems of finite sets, Quart. J. Math. Oxford Ser. (2) 12 (1961), 313–320, carded at Intersection theorems for systems of finite sets. Erdős's 1965 paper quotes this case of the problem as its equation (3), and Huang, Loh and Sudakov note that the case of two disjoint edges is equivalent to the Erdős–Ko–Rado theorem. The site's commentary attaches the theorem to $r=2$ in a parenthesis; the theorem gives the case $k=2$ for every $r$.

Covers. The case $k=2$ for every $r\ge3$ and every $n\ge2r$, the whole of the corrected Statement at $k=2$. It says nothing about $k\ge3$.

Depends on. No page of this wiki.

Acceptance. Refereed: the paper appeared in the Quarterly Journal of Mathematics, Oxford Second Series, in 1961 (volume 12, issue 1); the record gives only the year, so the page is dated to its first day. The site labels the problem FALSIFIABLE, an open label, so its commentary is not acceptance and no reviewed is listed. Nothing here rests on this project's own review.