TITLE: Tricky (extremal?) combinatorics problem
QUESTION [8 upvotes]: 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!

REPLY [2 votes]: 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