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For given $n$ and $k$, I want to form subsets with $k$ elements each from a base set of $n$ elements such that each pair of different elements $(e_i,e_j)$ is contained in the same subset equally often. As this is part of larger project in which I will have to do some computations for each subset I would like to find a solution with as few subsets as possible.

Some solutions for small n and k may help to illustrate the problem and what kind of solution I am searching for:

A trivial solution for all n and $n\ge k \ge2$ is to take all $\binom{n}{k}$ subsets. For this solution we need $\binom{n}{k}$ subsets, but all pairs of elements occur equally often, namely in $\binom{n-2}{k-2}$ subsets.

For $k=2$ this is also the best we can do, as there are $\binom{n}{2}$ pairs and we only cover one pair for each subset with two elements.

As a very concrete example for $n=4$ and $k=3$ with elements 1,2,3,4 we have to do the following 4 subsets: 123, 124, 134, 234 in which each pair occurs twice.

The smallest number, where I am currently unsure of the answer are $n=6$ and $k=3$. Is there a solution which requires only 10 instead of 20 subsets?

But how does this look in general? Are there any solutions which need less than $\binom{n}{k}$ subsets?

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    $\begingroup$ For $n=6$, $k=3$, I think $123, 145, 346, 256, 126, 146, 135, 356, 245, 234$ works. $\endgroup$ – Jaap Scherphuis Aug 28 '20 at 14:52
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A family of $k$-subsets where each pair occurs $\lambda$ times is called a $2$-design or BIBD or more specifically a $(n, k, \lambda)$-design. They are a special case (and it seems historically the first case?) of block designs.

If I understand correctly, you seem to be seeking, for given $n, k$, the smallest $\lambda$ (which would of course determine the smallest such family). This question is not specifically addressed by that wikipedia article, but since the field is almost a century old hopefully you can find the results somewhere else now that you have some search terms. However, I would not be surprised if you cannot find a general formula for arbitrary $(n, k)$.

Good luck!

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  • $\begingroup$ Thanks, This is indeed helpful as a guide for where to look! The btw is wrong though, all pairs which are in separate subsets do not occur... $\endgroup$ – Xenon Aug 30 '20 at 13:21
  • $\begingroup$ Hahaha, of course, what was I thinking! :D I deleted that part now. Sorry! $\endgroup$ – antkam Aug 30 '20 at 16:55

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