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Let $X$ be a set with $n$ elements. I would like to know the maximum number of subsets of $X$ such that the number of elements in the symmetric difference between any two of these subsets is at most $k$.

For example if $k=n$, then the answer is $2^n$.

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  • $\begingroup$ Represent each subset as a vector $\{0,1\}^n$ with $1$ at position $i$ meaning the subset contains the $i$th element of $X$. The symmetric difference between two subsets is the Hamming difference between these two vectors. So your question is equivalent to asking about length-$n$ binary error-correcting codes, except instead of having a minimum distance ("$d$") you have a maximum distance ("$k$"). $\endgroup$ Commented Sep 17, 2015 at 13:14
  • $\begingroup$ Maybe the answer should be the size of the Hamming ball of radius $k/2$, since one can take all vectors in this ball and be assured that they are of distance at most $k$ from each other, but with any more vectors this is not true...? $\endgroup$ Commented Sep 17, 2015 at 13:17

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The maximum is attained at the family of all subsets of cardinality $\leq k/2$ if $k$ is even, and of all subsets whose intersection with $X\setminus\{x\}$ is of cardinatliy at most $(k-1)/2$ if $k$ is odd (here $x$ is some fixed element in $X$). This is proved by Kleitman in D. Kleitman, On a combinatorial conjecture of Erdös, J. Combin. Theory, Vol. 1, 209–214, at least for even values of $k$. P. Frankl attributes the full version to Kleitman (see Theorem 5.3 in P, Frankl, Extremal Set Systems, in R.L. Graham, M. Grötschel, L. Lovász (eds.), Handbook of Combinatorics, Vol. 2, Elsevier, 1995; pp. 1293--1330), but I am not sure whether Kleitman has actually claimed this. Anyway, the full version is proved in the cited Frankl's review.

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