Tricky (extremal?) combinatorics problem

Apologies for being unsure the best way to express this problem.

I have 9 tables with 4 students at each table. I want to re-seat all students so no two students who have sat together ever sit together again. What’s the maximum possible number of re-seatings?

Generally, given a set S of students partitioned into n disjoint subsets of k elements each, what’s the maximum number of times I can permute the elements of S such that no two elements are members of the same subsets more than once. (I think that’s what I want to be asking).

If anyone can point me in the right direction or provide illumination, it would be greatly appreciated!

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The keyword here is block design (and resolvable block designs).

Update: The maximum number of rounds is between $9$ and $11$.

  • Person 1 sits with $3$ distinct people in each round, so we can have at most $\lfloor (36-1)/3 \rfloor=11$ rounds. (We could similarly argue that there are $\binom{36}{2}$ pairs and pairs are used $9\binom{4}{2}$ at a time, giving $\leq 11$ possible rounds.)

  • $9$ rounds is possible, and can be constructed as follows:

    • Step 1: Take three mutually orthogonal Latin squares of order $9$ (which exist; in fact, a finite field construction gives $8$-$\mathrm{MOLS}(9)$ (ref.)). Call these $L_1,L_2,L_3$ and assume $L_1$ uses the symbol $\{1,\ldots,9\}$, $L_2$ uses the symbol $\{10,\ldots,18\}$ and $L_3$ uses the symbol $\{19,\ldots,27\}$.

    • Step 2: Take the “union” of these matrices to form a $9 \times 9$ matrix in which each cell $(i,j)$ contains $\{L_1[i,j],L_2[i,j],L_3[i,j]\}$.

    • Step 3: Add symbol $28$ to the sets in the first column, $29$ to the sets in the second column, and so on, up to $36$ in the last column.

    We can check there are no pairs that occur twice case-by-case: if $x$ and $y$ (with $x \neq y$) occurs in two sets, then either (a) $(x,y)$ occurs in some $(L_i,L_j)$ twice, contradicting the orthogonal property, or (b) both $y$s occur in the same column, either contradicting that the $L_i$s are Latin squares, or contradicting $x \neq y$.

As a specific example:


It’s not clear to me if more rounds than this would be possible.

I wrote some code that found a bajillion random $8$-round examples, e.g.:


None of these $8$-round examples were extendable.

The random examples made it look like finding a $9$-round example this way would be difficult — there were very few possible table combinations remaining, let alone finding $9$ simultaneous table combinations including every person. In the above example, for instance, no valid table can be formed with person $6$.

This is the social golfer problem. Here’s the maximal solution for 9 groups of 4.

day 1:  BCNP    HRlm    Dkor    Fhnp    AJaj    KLEG    QIcd    Mbfi    Oqeg
day 2:  CDOQ    IAmn    Elpa    Gioq    BKbk    LMFH    RJde    Ncgj    Prfh
day 3:  DEPR    JBno    Fmqb    Hjpr    CLcl    MNGI    AKef    Odhk    Qagi
day 4:  EFQA    KCop    Gnrc    Ikqa    DMdm    NOHJ    BLfg    Peil    Rbhj
day 5:  FGRB    LDpq    Hoad    Jlrb    ENen    OPIK    CMgh    Qfjm    Acik
day 6:  GHAC    MEqr    Ipbe    Kmac    FOfo    PQJL    DNhi    Rgkn    Bdjl
day 7:  HIBD    NFra    Jqcf    Lnbd    GPgp    QRKM    EOij    Ahlo    Cekm
day 8:  IJCE    OGab    Krdg    Moce    HQhq    RALN    FPjk    Bimp    Dfln
day 9:  JKDF    PHbc    Laeh    Npdf    IRir    ABMO    GQkl    Cjnq    Egmo
day 10: BEch    DGej    FIgl    HKin    JMkp    LOmr    NQob    PAqd    RCaf
day 11: CFdi    EHfk    GJhm    ILjo    KNlq    MPna    ORpc    QBre    ADbg