Number of Partitions of n into 4 parts equals the number of partitions of 3n into 4 parts of size at most...












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Let $ngeq 4$. Prove that the number of partitions of $n$ into 4 parts equals the number of partitions of $3n$ into 4 parts of size at most $n-1$.



I am stuck on this problem but I suspect I need to establish a bijection by possibly looking at the conjugate of the Ferrer's diagram. Thanks for any help.






combinatorics integer-partitions







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$endgroup$












  • $begingroup$
    What have you tried so far?
    $endgroup$
    – platty
    Dec 5 '18 at 17:42






  • 1




    $begingroup$
    You have asked enough questions here to know that the right way to ask for help is to edit this one to show us what you tried and where you are stuck.
    $endgroup$
    – Ethan Bolker
    Dec 5 '18 at 17:43






  • 4




    $begingroup$
    Hint: A quadruple $(x_1,x_2,x_3,x_4)inmathbb{Z}_{>0}^4$ with $x_1leq x_2leq x_3leq x_4$ partitions $n$ if and only if $(n-x_4,n-x_3,n-x_2,n-x_1)$ partitions $3n$.
    $endgroup$
    – Batominovski
    Dec 5 '18 at 17:46


















0












$begingroup$


Let $ngeq 4$. Prove that the number of partitions of $n$ into 4 parts equals the number of partitions of $3n$ into 4 parts of size at most $n-1$.



I am stuck on this problem but I suspect I need to establish a bijection by possibly looking at the conjugate of the Ferrer's diagram. Thanks for any help.






combinatorics integer-partitions







share|cite|improve this question











$endgroup$












  • $begingroup$
    What have you tried so far?
    $endgroup$
    – platty
    Dec 5 '18 at 17:42






  • 1




    $begingroup$
    You have asked enough questions here to know that the right way to ask for help is to edit this one to show us what you tried and where you are stuck.
    $endgroup$
    – Ethan Bolker
    Dec 5 '18 at 17:43






  • 4




    $begingroup$
    Hint: A quadruple $(x_1,x_2,x_3,x_4)inmathbb{Z}_{>0}^4$ with $x_1leq x_2leq x_3leq x_4$ partitions $n$ if and only if $(n-x_4,n-x_3,n-x_2,n-x_1)$ partitions $3n$.
    $endgroup$
    – Batominovski
    Dec 5 '18 at 17:46
















0












0








0


2



$begingroup$


Let $ngeq 4$. Prove that the number of partitions of $n$ into 4 parts equals the number of partitions of $3n$ into 4 parts of size at most $n-1$.



I am stuck on this problem but I suspect I need to establish a bijection by possibly looking at the conjugate of the Ferrer's diagram. Thanks for any help.






combinatorics integer-partitions







share|cite|improve this question











$endgroup$




Let $ngeq 4$. Prove that the number of partitions of $n$ into 4 parts equals the number of partitions of $3n$ into 4 parts of size at most $n-1$.



I am stuck on this problem but I suspect I need to establish a bijection by possibly looking at the conjugate of the Ferrer's diagram. Thanks for any help.






combinatorics integer-partitions




combinatorics integer-partitions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 17:49







TNT

















asked Dec 5 '18 at 17:39









TNTTNT

596




596












  • $begingroup$
    What have you tried so far?
    $endgroup$
    – platty
    Dec 5 '18 at 17:42






  • 1




    $begingroup$
    You have asked enough questions here to know that the right way to ask for help is to edit this one to show us what you tried and where you are stuck.
    $endgroup$
    – Ethan Bolker
    Dec 5 '18 at 17:43






  • 4




    $begingroup$
    Hint: A quadruple $(x_1,x_2,x_3,x_4)inmathbb{Z}_{>0}^4$ with $x_1leq x_2leq x_3leq x_4$ partitions $n$ if and only if $(n-x_4,n-x_3,n-x_2,n-x_1)$ partitions $3n$.
    $endgroup$
    – Batominovski
    Dec 5 '18 at 17:46




















  • $begingroup$
    What have you tried so far?
    $endgroup$
    – platty
    Dec 5 '18 at 17:42






  • 1




    $begingroup$
    You have asked enough questions here to know that the right way to ask for help is to edit this one to show us what you tried and where you are stuck.
    $endgroup$
    – Ethan Bolker
    Dec 5 '18 at 17:43






  • 4




    $begingroup$
    Hint: A quadruple $(x_1,x_2,x_3,x_4)inmathbb{Z}_{>0}^4$ with $x_1leq x_2leq x_3leq x_4$ partitions $n$ if and only if $(n-x_4,n-x_3,n-x_2,n-x_1)$ partitions $3n$.
    $endgroup$
    – Batominovski
    Dec 5 '18 at 17:46


















$begingroup$
What have you tried so far?
$endgroup$
– platty
Dec 5 '18 at 17:42




$begingroup$
What have you tried so far?
$endgroup$
– platty
Dec 5 '18 at 17:42




1




1




$begingroup$
You have asked enough questions here to know that the right way to ask for help is to edit this one to show us what you tried and where you are stuck.
$endgroup$
– Ethan Bolker
Dec 5 '18 at 17:43




$begingroup$
You have asked enough questions here to know that the right way to ask for help is to edit this one to show us what you tried and where you are stuck.
$endgroup$
– Ethan Bolker
Dec 5 '18 at 17:43




4




4




$begingroup$
Hint: A quadruple $(x_1,x_2,x_3,x_4)inmathbb{Z}_{>0}^4$ with $x_1leq x_2leq x_3leq x_4$ partitions $n$ if and only if $(n-x_4,n-x_3,n-x_2,n-x_1)$ partitions $3n$.
$endgroup$
– Batominovski
Dec 5 '18 at 17:46






$begingroup$
Hint: A quadruple $(x_1,x_2,x_3,x_4)inmathbb{Z}_{>0}^4$ with $x_1leq x_2leq x_3leq x_4$ partitions $n$ if and only if $(n-x_4,n-x_3,n-x_2,n-x_1)$ partitions $3n$.
$endgroup$
– Batominovski
Dec 5 '18 at 17:46












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Proof: Let $ngeq 4$. The number of partitions of $n$ into 4 parts is a solution to the system $x_1+x_2+x_3+x_4=n$ for $x_1geq x_2geq x_3geq x_4geq 1$. Let $y_1=n-x_4$, $y_2=n-x_3$, $y_3=n-x_2$, and $y_4=n-x_1$. We know that $x_1geq x_2geq x_3geq x_4geq 1$ which implies $-x_1leq -x_2leq -x_3leq -x_4leq -1$ and thus $n-x_1leq n-x_2leq n-x_3leq n-x_4leq n-1$. So, $1leq y_4leq y_3leq y_2leq y_1leq n-1$. Hence $y_1+y_2+y_3+y_4=4n-(x_1+x_2+x_3+x_4)=4n-n=3n$ which yields a solution to the number of partitions of $3n$ into 4 parts of size at most $n-1$. Thus there is a bijection between the system $y_1+y_2+y_3+y_4=3n$ for $n-1geq y_4geq y_3geq y_2geq y_1geq 1$ and the system $x_1+x_2+x_3+x_4=n$ for $x_1geq x_2geq x_3geq x_4geq 1$. Therefore, the number of partitions of $n$
into 4 parts equals the number of partitions of $3n$ into 4 parts of
size at most $n-1$.






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    $begingroup$

    Proof: Let $ngeq 4$. The number of partitions of $n$ into 4 parts is a solution to the system $x_1+x_2+x_3+x_4=n$ for $x_1geq x_2geq x_3geq x_4geq 1$. Let $y_1=n-x_4$, $y_2=n-x_3$, $y_3=n-x_2$, and $y_4=n-x_1$. We know that $x_1geq x_2geq x_3geq x_4geq 1$ which implies $-x_1leq -x_2leq -x_3leq -x_4leq -1$ and thus $n-x_1leq n-x_2leq n-x_3leq n-x_4leq n-1$. So, $1leq y_4leq y_3leq y_2leq y_1leq n-1$. Hence $y_1+y_2+y_3+y_4=4n-(x_1+x_2+x_3+x_4)=4n-n=3n$ which yields a solution to the number of partitions of $3n$ into 4 parts of size at most $n-1$. Thus there is a bijection between the system $y_1+y_2+y_3+y_4=3n$ for $n-1geq y_4geq y_3geq y_2geq y_1geq 1$ and the system $x_1+x_2+x_3+x_4=n$ for $x_1geq x_2geq x_3geq x_4geq 1$. Therefore, the number of partitions of $n$
    into 4 parts equals the number of partitions of $3n$ into 4 parts of
    size at most $n-1$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Proof: Let $ngeq 4$. The number of partitions of $n$ into 4 parts is a solution to the system $x_1+x_2+x_3+x_4=n$ for $x_1geq x_2geq x_3geq x_4geq 1$. Let $y_1=n-x_4$, $y_2=n-x_3$, $y_3=n-x_2$, and $y_4=n-x_1$. We know that $x_1geq x_2geq x_3geq x_4geq 1$ which implies $-x_1leq -x_2leq -x_3leq -x_4leq -1$ and thus $n-x_1leq n-x_2leq n-x_3leq n-x_4leq n-1$. So, $1leq y_4leq y_3leq y_2leq y_1leq n-1$. Hence $y_1+y_2+y_3+y_4=4n-(x_1+x_2+x_3+x_4)=4n-n=3n$ which yields a solution to the number of partitions of $3n$ into 4 parts of size at most $n-1$. Thus there is a bijection between the system $y_1+y_2+y_3+y_4=3n$ for $n-1geq y_4geq y_3geq y_2geq y_1geq 1$ and the system $x_1+x_2+x_3+x_4=n$ for $x_1geq x_2geq x_3geq x_4geq 1$. Therefore, the number of partitions of $n$
      into 4 parts equals the number of partitions of $3n$ into 4 parts of
      size at most $n-1$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Proof: Let $ngeq 4$. The number of partitions of $n$ into 4 parts is a solution to the system $x_1+x_2+x_3+x_4=n$ for $x_1geq x_2geq x_3geq x_4geq 1$. Let $y_1=n-x_4$, $y_2=n-x_3$, $y_3=n-x_2$, and $y_4=n-x_1$. We know that $x_1geq x_2geq x_3geq x_4geq 1$ which implies $-x_1leq -x_2leq -x_3leq -x_4leq -1$ and thus $n-x_1leq n-x_2leq n-x_3leq n-x_4leq n-1$. So, $1leq y_4leq y_3leq y_2leq y_1leq n-1$. Hence $y_1+y_2+y_3+y_4=4n-(x_1+x_2+x_3+x_4)=4n-n=3n$ which yields a solution to the number of partitions of $3n$ into 4 parts of size at most $n-1$. Thus there is a bijection between the system $y_1+y_2+y_3+y_4=3n$ for $n-1geq y_4geq y_3geq y_2geq y_1geq 1$ and the system $x_1+x_2+x_3+x_4=n$ for $x_1geq x_2geq x_3geq x_4geq 1$. Therefore, the number of partitions of $n$
        into 4 parts equals the number of partitions of $3n$ into 4 parts of
        size at most $n-1$.






        share|cite|improve this answer









        $endgroup$



        Proof: Let $ngeq 4$. The number of partitions of $n$ into 4 parts is a solution to the system $x_1+x_2+x_3+x_4=n$ for $x_1geq x_2geq x_3geq x_4geq 1$. Let $y_1=n-x_4$, $y_2=n-x_3$, $y_3=n-x_2$, and $y_4=n-x_1$. We know that $x_1geq x_2geq x_3geq x_4geq 1$ which implies $-x_1leq -x_2leq -x_3leq -x_4leq -1$ and thus $n-x_1leq n-x_2leq n-x_3leq n-x_4leq n-1$. So, $1leq y_4leq y_3leq y_2leq y_1leq n-1$. Hence $y_1+y_2+y_3+y_4=4n-(x_1+x_2+x_3+x_4)=4n-n=3n$ which yields a solution to the number of partitions of $3n$ into 4 parts of size at most $n-1$. Thus there is a bijection between the system $y_1+y_2+y_3+y_4=3n$ for $n-1geq y_4geq y_3geq y_2geq y_1geq 1$ and the system $x_1+x_2+x_3+x_4=n$ for $x_1geq x_2geq x_3geq x_4geq 1$. Therefore, the number of partitions of $n$
        into 4 parts equals the number of partitions of $3n$ into 4 parts of
        size at most $n-1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 0:48









        TNTTNT

        596




        596






























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