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Is there a nice way to partition the edges of the complete 5-uniform hypergraph on 11 vertices into 7 copies of the Steiner system S(4,5,11)? If this is obvious or elementary, I apologize in advance.

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    $\begingroup$ By conjugating by random elements of $S_{11}$, it is easy to find pairs of disjoint Steiner systems $S(4,5,11)$, but I failed to find any triples of mutually disjoint systems, so the answer might be no. Of course that does not rule out the possibility of there being some clever way of doing the construction which you would not find using a random search. $\endgroup$ – Derek Holt Jun 8 '14 at 16:57
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Unfortunately, no. It is known that the maximum number of mutually disjoint $S(4,5,11)$s on the same point set is $2$. Any such pair are always isomorphic. So, you can't find $7$ disjoint copies of an $S(4,5,11)$ in the complete $5$-uniform hypergraph on $11$ vertices (or partition it into copies); you can find only two of them at most.

E. S. Kramer, D. M. Mesner, Intersections among Steiner systems, J. Combin. Theory, Ser. A, 16 (1974), 273–285

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