Families of ordered set partitions with disjoint blocks The 2019 Stack Overflow Developer Survey Results Are InDeligne-Simpson problem in the symmetric groupProcreation with several gendersI am searching for the name of a partition (if it already exists)Existence problem for a generalisation of Latin squares (matrices with fixed row and column sets)Simple lower bounds for Bell numbers (number of set partitions)?Can a partition free family in $2^[n]$ always be enlarged to one of size $2^n-1$?Looking for N-dimensional spheres in the configuration space of the colorful Tverberg problemBalanced partitions of vector setsCan we cover a set by a particular family of sets?genus zero permutation and noncrossing partition

Families of ordered set partitions with disjoint blocks



The 2019 Stack Overflow Developer Survey Results Are InDeligne-Simpson problem in the symmetric groupProcreation with several gendersI am searching for the name of a partition (if it already exists)Existence problem for a generalisation of Latin squares (matrices with fixed row and column sets)Simple lower bounds for Bell numbers (number of set partitions)?Can a partition free family in $2^[n]$ always be enlarged to one of size $2^n-1$?Looking for N-dimensional spheres in the configuration space of the colorful Tverberg problemBalanced partitions of vector setsCan we cover a set by a particular family of sets?genus zero permutation and noncrossing partition










2












$begingroup$


Let $C_1,dots, C_m$ be a family of ordered set partitions of $[n]$ with exactly $k$ blocks.



Write $C_i = B_i1, dots, B_ik$ for $i=1,dots, m$ where $B_ij$ are the blocks of the ordered set partition $C_i$.



Suppose this family also has the property that for each $j=1,dots, k$



$$B_1j cup cdots cup B_mj$$



is also a partition of $[n]$



Can one determine the maximal number of members in such a family $m$, or at least a decent upper bound on $m$?



Edit:



It might also be worth noting that if we take $k=n$, then $m=n$ since this would be equivalent to the existence of a latin square. I am in particular interested in the case $k=2$.






co.combinatorics partitions







share|cite|improve this question











$endgroup$
















    2












    $begingroup$


    Let $C_1,dots, C_m$ be a family of ordered set partitions of $[n]$ with exactly $k$ blocks.



    Write $C_i = B_i1, dots, B_ik$ for $i=1,dots, m$ where $B_ij$ are the blocks of the ordered set partition $C_i$.



    Suppose this family also has the property that for each $j=1,dots, k$



    $$B_1j cup cdots cup B_mj$$



    is also a partition of $[n]$



    Can one determine the maximal number of members in such a family $m$, or at least a decent upper bound on $m$?



    Edit:



    It might also be worth noting that if we take $k=n$, then $m=n$ since this would be equivalent to the existence of a latin square. I am in particular interested in the case $k=2$.






    co.combinatorics partitions







    share|cite|improve this question











    $endgroup$














      2












      2








      2





      $begingroup$


      Let $C_1,dots, C_m$ be a family of ordered set partitions of $[n]$ with exactly $k$ blocks.



      Write $C_i = B_i1, dots, B_ik$ for $i=1,dots, m$ where $B_ij$ are the blocks of the ordered set partition $C_i$.



      Suppose this family also has the property that for each $j=1,dots, k$



      $$B_1j cup cdots cup B_mj$$



      is also a partition of $[n]$



      Can one determine the maximal number of members in such a family $m$, or at least a decent upper bound on $m$?



      Edit:



      It might also be worth noting that if we take $k=n$, then $m=n$ since this would be equivalent to the existence of a latin square. I am in particular interested in the case $k=2$.






      co.combinatorics partitions







      share|cite|improve this question











      $endgroup$




      Let $C_1,dots, C_m$ be a family of ordered set partitions of $[n]$ with exactly $k$ blocks.



      Write $C_i = B_i1, dots, B_ik$ for $i=1,dots, m$ where $B_ij$ are the blocks of the ordered set partition $C_i$.



      Suppose this family also has the property that for each $j=1,dots, k$



      $$B_1j cup cdots cup B_mj$$



      is also a partition of $[n]$



      Can one determine the maximal number of members in such a family $m$, or at least a decent upper bound on $m$?



      Edit:



      It might also be worth noting that if we take $k=n$, then $m=n$ since this would be equivalent to the existence of a latin square. I am in particular interested in the case $k=2$.






      co.combinatorics partitions




      co.combinatorics partitions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 8 hours ago









      darij grinberg

      18.4k373188




      18.4k373188










      asked 10 hours ago









      user94267user94267

      1006




      1006




















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          We have $$mn=sum_isum_j |B_ij|=sum_jsum_i |B_ij|=kn,$$
          thus $m=k$.






          share|cite|improve this answer









          $endgroup$




















            3












            $begingroup$

            Answer: $m=k$.



            Put indeed your blocks $B_ij$ in a $mtimes k$ array and then "read" this array:



            -- row-wise: any element of $[n]$ appears then $m$ times.



            -- column-wise: any element of $[n]$ appears then $k$ times.






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              26 seconds slower than Fedor, but my answer is better, does not use multiplication :)
              $endgroup$
              – Teo Banica
              10 hours ago










            • $begingroup$
              you actually multiply 1 by $m$ and by $k$ :)
              $endgroup$
              – Fedor Petrov
              10 hours ago











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            2 Answers
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            active

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            2 Answers
            2






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            We have $$mn=sum_isum_j |B_ij|=sum_jsum_i |B_ij|=kn,$$
            thus $m=k$.






            share|cite|improve this answer









            $endgroup$

















              3












              $begingroup$

              We have $$mn=sum_isum_j |B_ij|=sum_jsum_i |B_ij|=kn,$$
              thus $m=k$.






              share|cite|improve this answer









              $endgroup$















                3












                3








                3





                $begingroup$

                We have $$mn=sum_isum_j |B_ij|=sum_jsum_i |B_ij|=kn,$$
                thus $m=k$.






                share|cite|improve this answer









                $endgroup$



                We have $$mn=sum_isum_j |B_ij|=sum_jsum_i |B_ij|=kn,$$
                thus $m=k$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 10 hours ago









                Fedor PetrovFedor Petrov

                52.1k6122239




                52.1k6122239





















                    3












                    $begingroup$

                    Answer: $m=k$.



                    Put indeed your blocks $B_ij$ in a $mtimes k$ array and then "read" this array:



                    -- row-wise: any element of $[n]$ appears then $m$ times.



                    -- column-wise: any element of $[n]$ appears then $k$ times.






                    share|cite|improve this answer









                    $endgroup$












                    • $begingroup$
                      26 seconds slower than Fedor, but my answer is better, does not use multiplication :)
                      $endgroup$
                      – Teo Banica
                      10 hours ago










                    • $begingroup$
                      you actually multiply 1 by $m$ and by $k$ :)
                      $endgroup$
                      – Fedor Petrov
                      10 hours ago















                    3












                    $begingroup$

                    Answer: $m=k$.



                    Put indeed your blocks $B_ij$ in a $mtimes k$ array and then "read" this array:



                    -- row-wise: any element of $[n]$ appears then $m$ times.



                    -- column-wise: any element of $[n]$ appears then $k$ times.






                    share|cite|improve this answer









                    $endgroup$












                    • $begingroup$
                      26 seconds slower than Fedor, but my answer is better, does not use multiplication :)
                      $endgroup$
                      – Teo Banica
                      10 hours ago










                    • $begingroup$
                      you actually multiply 1 by $m$ and by $k$ :)
                      $endgroup$
                      – Fedor Petrov
                      10 hours ago













                    3












                    3








                    3





                    $begingroup$

                    Answer: $m=k$.



                    Put indeed your blocks $B_ij$ in a $mtimes k$ array and then "read" this array:



                    -- row-wise: any element of $[n]$ appears then $m$ times.



                    -- column-wise: any element of $[n]$ appears then $k$ times.






                    share|cite|improve this answer









                    $endgroup$



                    Answer: $m=k$.



                    Put indeed your blocks $B_ij$ in a $mtimes k$ array and then "read" this array:



                    -- row-wise: any element of $[n]$ appears then $m$ times.



                    -- column-wise: any element of $[n]$ appears then $k$ times.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 10 hours ago









                    Teo BanicaTeo Banica

                    568528




                    568528











                    • $begingroup$
                      26 seconds slower than Fedor, but my answer is better, does not use multiplication :)
                      $endgroup$
                      – Teo Banica
                      10 hours ago










                    • $begingroup$
                      you actually multiply 1 by $m$ and by $k$ :)
                      $endgroup$
                      – Fedor Petrov
                      10 hours ago
















                    • $begingroup$
                      26 seconds slower than Fedor, but my answer is better, does not use multiplication :)
                      $endgroup$
                      – Teo Banica
                      10 hours ago










                    • $begingroup$
                      you actually multiply 1 by $m$ and by $k$ :)
                      $endgroup$
                      – Fedor Petrov
                      10 hours ago















                    $begingroup$
                    26 seconds slower than Fedor, but my answer is better, does not use multiplication :)
                    $endgroup$
                    – Teo Banica
                    10 hours ago




                    $begingroup$
                    26 seconds slower than Fedor, but my answer is better, does not use multiplication :)
                    $endgroup$
                    – Teo Banica
                    10 hours ago












                    $begingroup$
                    you actually multiply 1 by $m$ and by $k$ :)
                    $endgroup$
                    – Fedor Petrov
                    10 hours ago




                    $begingroup$
                    you actually multiply 1 by $m$ and by $k$ :)
                    $endgroup$
                    – Fedor Petrov
                    10 hours ago

















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