3

Let $S$ be a set of elements, e.g. $S=\{ 1, 2, 3, 4, 5, 6 \}$ and $M \subset S$ where $|M| = n$. How many subsets $M$ of size $n$ are at least necessary such that each $s \in S$ is at least once in a subset $M$ with each other element of $S$? For example, let $n = 4$, we need $3$ subsets $$ \{1,2,3,4\}, \{3,4,5,6\}, \{1,2,5,6\} $$ because $1$ is in a subset with each other element, $2$ as well, and so on.

Another example: Let $S = \{A, B, C, D, E, F\}$

A B C D E F
x x x
x x x
x x x
x x x
x x x
x x x

In this case, my best solutions requires $6$ subsets.

xndr
  • 83
  • How far have you gotten with this? – saulspatz Apr 13 '21 at 18:58
  • I know for the case $n = 2$ we can use $\binom{6}{2} =15$. – xndr Apr 13 '21 at 19:01
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    Clearly you need at least ${|S| \choose 2}/{n \choose 2}$ subsets, since there are ${|S| \choose 2}$ pairs that need to be covered and you get ${n \choose 2}$ pairs from each subset. In the case where $|S| = 6, n = 4$ you get ${6 \choose 2}/{4 \choose 2} = 15/6 = 2.5$, so 3 is the best possible, which you've acheived. – Michael Lugo Apr 13 '21 at 19:11
  • But for $|S| = 6$ and $n = 3$ I need at least 7, or am I wrong? – xndr Apr 13 '21 at 19:27
  • Ok, 7 is wrong, I found a solution with 6 subsets, but I can't find a solution, where I only need 5 subsets. – xndr Apr 13 '21 at 20:00

0 Answers0