Bijection between $mathsf{Sh}(p+1, q)$ and $mathsf{Sh}(p, q)sqcup mathsf{Sh}(p-1, q+1)$?












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$begingroup$


Let $p$ and $q$ be two positive integers. Define $$mathsf{Sh}(p, q):={sigmainmathsf{S}_{p+q}: sigma(1)<ldots<sigma(p) textrm{and} sigma(p+1)<ldots<sigma(p+q)},$$ where $mathsf{S}_{p+q}$ is the set of permutations of the first $p+q$ integers, i.e, bijections of ${1, ldots, p+q}$. An element of $mathsf{Sh}(p, q)$ is called a $(p, q)$-shuffle.



It can be seen that $$#mathsf{Sh}(p, q)=binom{p+q}{p}.$$



In particular, we see that $$mathsf{Sh}(p+1, q)=#mathsf{Sh}(p, q)+#mathsf{Sh}(p+1, q-1),$$ and therefore $$#mathsf{Sh}(p+1, q)quad textrm{and}quad mathsf{Sh}(p, q)sqcup mathsf{Sh}(p+1, q-1)$$
are in bijection. How to construct an explicit bijection between those sets?



Thanks.






combinatorics







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    0












    $begingroup$


    Let $p$ and $q$ be two positive integers. Define $$mathsf{Sh}(p, q):={sigmainmathsf{S}_{p+q}: sigma(1)<ldots<sigma(p) textrm{and} sigma(p+1)<ldots<sigma(p+q)},$$ where $mathsf{S}_{p+q}$ is the set of permutations of the first $p+q$ integers, i.e, bijections of ${1, ldots, p+q}$. An element of $mathsf{Sh}(p, q)$ is called a $(p, q)$-shuffle.



    It can be seen that $$#mathsf{Sh}(p, q)=binom{p+q}{p}.$$



    In particular, we see that $$mathsf{Sh}(p+1, q)=#mathsf{Sh}(p, q)+#mathsf{Sh}(p+1, q-1),$$ and therefore $$#mathsf{Sh}(p+1, q)quad textrm{and}quad mathsf{Sh}(p, q)sqcup mathsf{Sh}(p+1, q-1)$$
    are in bijection. How to construct an explicit bijection between those sets?



    Thanks.






    combinatorics







    share|cite|improve this question











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      0





      $begingroup$


      Let $p$ and $q$ be two positive integers. Define $$mathsf{Sh}(p, q):={sigmainmathsf{S}_{p+q}: sigma(1)<ldots<sigma(p) textrm{and} sigma(p+1)<ldots<sigma(p+q)},$$ where $mathsf{S}_{p+q}$ is the set of permutations of the first $p+q$ integers, i.e, bijections of ${1, ldots, p+q}$. An element of $mathsf{Sh}(p, q)$ is called a $(p, q)$-shuffle.



      It can be seen that $$#mathsf{Sh}(p, q)=binom{p+q}{p}.$$



      In particular, we see that $$mathsf{Sh}(p+1, q)=#mathsf{Sh}(p, q)+#mathsf{Sh}(p+1, q-1),$$ and therefore $$#mathsf{Sh}(p+1, q)quad textrm{and}quad mathsf{Sh}(p, q)sqcup mathsf{Sh}(p+1, q-1)$$
      are in bijection. How to construct an explicit bijection between those sets?



      Thanks.






      combinatorics







      share|cite|improve this question











      $endgroup$




      Let $p$ and $q$ be two positive integers. Define $$mathsf{Sh}(p, q):={sigmainmathsf{S}_{p+q}: sigma(1)<ldots<sigma(p) textrm{and} sigma(p+1)<ldots<sigma(p+q)},$$ where $mathsf{S}_{p+q}$ is the set of permutations of the first $p+q$ integers, i.e, bijections of ${1, ldots, p+q}$. An element of $mathsf{Sh}(p, q)$ is called a $(p, q)$-shuffle.



      It can be seen that $$#mathsf{Sh}(p, q)=binom{p+q}{p}.$$



      In particular, we see that $$mathsf{Sh}(p+1, q)=#mathsf{Sh}(p, q)+#mathsf{Sh}(p+1, q-1),$$ and therefore $$#mathsf{Sh}(p+1, q)quad textrm{and}quad mathsf{Sh}(p, q)sqcup mathsf{Sh}(p+1, q-1)$$
      are in bijection. How to construct an explicit bijection between those sets?



      Thanks.






      combinatorics




      combinatorics






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      edited Dec 4 '18 at 23:11







      PtF

















      asked Jun 3 '17 at 23:15









      PtFPtF

      3,94921733




      3,94921733






















          2 Answers
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          $begingroup$

          As stated, you have a wrong identity. Notice that $Sh(3+1,2)=binom{6}{4}=15,Sh(3,2)=binom{5}{3}=10$ and $Sh(2,3)=binom{5}{2}=10$ but $15neq 10+10.$


          I am going to make the assumption that you wanted to use Pascal's recursion, in which case the correct identity is $$|Sh(p,q+1)|=|Sh(p,q)|+|Sh(p-1,q+1)|.$$



          In order to prove the later, let $sigma in Sh(p,q+1),$ then there are $2$ options:

          $bullet $ $sigma ^{-1}(p+q+1)<=p$: if you remove $p+q+1$ you end up with a permutation in $Sh(pcolor{red}{-1},q+1).$


          $bullet$ $sigma ^{-1}(p+q+1)>p$:if you remove $p+q+1$ you end up with a permutation in $Sh(p,q+1color{red}{-1})=Sh(p,q).$

          It should be clear the bijection from here.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Sorry, I mistyped. Thanks for your answer =)
            $endgroup$
            – PtF
            Jun 4 '17 at 12:38










          • $begingroup$
            Sure, glad to help.
            $endgroup$
            – Phicar
            Jun 4 '17 at 16:20



















          0












          $begingroup$

          I wrote down the details so I'm going to share just in case someone else needs it someday.



          The first thing to be noticed is that $sigma^{-1}({1})={1, p+2}$, hence:



          $$mathsf{Sh}(p+1, q)={sigmain mathsf{Sh}(p+1, q): sigma(1)=1}sqcup {sigmainmathsf{Sh}(p+1, q): sigma(p+2)=1}.$$
          This allows us to define the bijection easily setting:



          $$mathsf{Sh}(p, q)longrightarrow mathsf{Sh}(p+1, q)\ sigmalongmapsto left(begin{array}{ccccccc}1 & 2& ldots & p+2 & ldots &p+q+1\ 1 &sigma(1)+1 & ldots &sigma(p+1)+1 & ldots & sigma(p+q)+1 end{array}right)$$
          and



          $$ mathsf{Sh}(p+1, q-1)longrightarrow mathsf{Sh}(p+1, q)\ sigmalongmapsto left(begin{array}{ccccccc}1 & ldots &p+1 & p+2 & p+3& ldots & p+q+1\ sigma(1)+1 & ldots & sigma(p+1)+1 & 1 & sigma(p+2)+1 & ldots &sigma(p+q)+1 end{array}right).$$ The inverse should be clear from the definitions.






          share|cite|improve this answer









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            2 Answers
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            $begingroup$

            As stated, you have a wrong identity. Notice that $Sh(3+1,2)=binom{6}{4}=15,Sh(3,2)=binom{5}{3}=10$ and $Sh(2,3)=binom{5}{2}=10$ but $15neq 10+10.$


            I am going to make the assumption that you wanted to use Pascal's recursion, in which case the correct identity is $$|Sh(p,q+1)|=|Sh(p,q)|+|Sh(p-1,q+1)|.$$



            In order to prove the later, let $sigma in Sh(p,q+1),$ then there are $2$ options:

            $bullet $ $sigma ^{-1}(p+q+1)<=p$: if you remove $p+q+1$ you end up with a permutation in $Sh(pcolor{red}{-1},q+1).$


            $bullet$ $sigma ^{-1}(p+q+1)>p$:if you remove $p+q+1$ you end up with a permutation in $Sh(p,q+1color{red}{-1})=Sh(p,q).$

            It should be clear the bijection from here.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Sorry, I mistyped. Thanks for your answer =)
              $endgroup$
              – PtF
              Jun 4 '17 at 12:38










            • $begingroup$
              Sure, glad to help.
              $endgroup$
              – Phicar
              Jun 4 '17 at 16:20
















            2












            $begingroup$

            As stated, you have a wrong identity. Notice that $Sh(3+1,2)=binom{6}{4}=15,Sh(3,2)=binom{5}{3}=10$ and $Sh(2,3)=binom{5}{2}=10$ but $15neq 10+10.$


            I am going to make the assumption that you wanted to use Pascal's recursion, in which case the correct identity is $$|Sh(p,q+1)|=|Sh(p,q)|+|Sh(p-1,q+1)|.$$



            In order to prove the later, let $sigma in Sh(p,q+1),$ then there are $2$ options:

            $bullet $ $sigma ^{-1}(p+q+1)<=p$: if you remove $p+q+1$ you end up with a permutation in $Sh(pcolor{red}{-1},q+1).$


            $bullet$ $sigma ^{-1}(p+q+1)>p$:if you remove $p+q+1$ you end up with a permutation in $Sh(p,q+1color{red}{-1})=Sh(p,q).$

            It should be clear the bijection from here.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Sorry, I mistyped. Thanks for your answer =)
              $endgroup$
              – PtF
              Jun 4 '17 at 12:38










            • $begingroup$
              Sure, glad to help.
              $endgroup$
              – Phicar
              Jun 4 '17 at 16:20














            2












            2








            2





            $begingroup$

            As stated, you have a wrong identity. Notice that $Sh(3+1,2)=binom{6}{4}=15,Sh(3,2)=binom{5}{3}=10$ and $Sh(2,3)=binom{5}{2}=10$ but $15neq 10+10.$


            I am going to make the assumption that you wanted to use Pascal's recursion, in which case the correct identity is $$|Sh(p,q+1)|=|Sh(p,q)|+|Sh(p-1,q+1)|.$$



            In order to prove the later, let $sigma in Sh(p,q+1),$ then there are $2$ options:

            $bullet $ $sigma ^{-1}(p+q+1)<=p$: if you remove $p+q+1$ you end up with a permutation in $Sh(pcolor{red}{-1},q+1).$


            $bullet$ $sigma ^{-1}(p+q+1)>p$:if you remove $p+q+1$ you end up with a permutation in $Sh(p,q+1color{red}{-1})=Sh(p,q).$

            It should be clear the bijection from here.






            share|cite|improve this answer









            $endgroup$



            As stated, you have a wrong identity. Notice that $Sh(3+1,2)=binom{6}{4}=15,Sh(3,2)=binom{5}{3}=10$ and $Sh(2,3)=binom{5}{2}=10$ but $15neq 10+10.$


            I am going to make the assumption that you wanted to use Pascal's recursion, in which case the correct identity is $$|Sh(p,q+1)|=|Sh(p,q)|+|Sh(p-1,q+1)|.$$



            In order to prove the later, let $sigma in Sh(p,q+1),$ then there are $2$ options:

            $bullet $ $sigma ^{-1}(p+q+1)<=p$: if you remove $p+q+1$ you end up with a permutation in $Sh(pcolor{red}{-1},q+1).$


            $bullet$ $sigma ^{-1}(p+q+1)>p$:if you remove $p+q+1$ you end up with a permutation in $Sh(p,q+1color{red}{-1})=Sh(p,q).$

            It should be clear the bijection from here.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jun 4 '17 at 5:12









            PhicarPhicar

            2,6251915




            2,6251915












            • $begingroup$
              Sorry, I mistyped. Thanks for your answer =)
              $endgroup$
              – PtF
              Jun 4 '17 at 12:38










            • $begingroup$
              Sure, glad to help.
              $endgroup$
              – Phicar
              Jun 4 '17 at 16:20


















            • $begingroup$
              Sorry, I mistyped. Thanks for your answer =)
              $endgroup$
              – PtF
              Jun 4 '17 at 12:38










            • $begingroup$
              Sure, glad to help.
              $endgroup$
              – Phicar
              Jun 4 '17 at 16:20
















            $begingroup$
            Sorry, I mistyped. Thanks for your answer =)
            $endgroup$
            – PtF
            Jun 4 '17 at 12:38




            $begingroup$
            Sorry, I mistyped. Thanks for your answer =)
            $endgroup$
            – PtF
            Jun 4 '17 at 12:38












            $begingroup$
            Sure, glad to help.
            $endgroup$
            – Phicar
            Jun 4 '17 at 16:20




            $begingroup$
            Sure, glad to help.
            $endgroup$
            – Phicar
            Jun 4 '17 at 16:20











            0












            $begingroup$

            I wrote down the details so I'm going to share just in case someone else needs it someday.



            The first thing to be noticed is that $sigma^{-1}({1})={1, p+2}$, hence:



            $$mathsf{Sh}(p+1, q)={sigmain mathsf{Sh}(p+1, q): sigma(1)=1}sqcup {sigmainmathsf{Sh}(p+1, q): sigma(p+2)=1}.$$
            This allows us to define the bijection easily setting:



            $$mathsf{Sh}(p, q)longrightarrow mathsf{Sh}(p+1, q)\ sigmalongmapsto left(begin{array}{ccccccc}1 & 2& ldots & p+2 & ldots &p+q+1\ 1 &sigma(1)+1 & ldots &sigma(p+1)+1 & ldots & sigma(p+q)+1 end{array}right)$$
            and



            $$ mathsf{Sh}(p+1, q-1)longrightarrow mathsf{Sh}(p+1, q)\ sigmalongmapsto left(begin{array}{ccccccc}1 & ldots &p+1 & p+2 & p+3& ldots & p+q+1\ sigma(1)+1 & ldots & sigma(p+1)+1 & 1 & sigma(p+2)+1 & ldots &sigma(p+q)+1 end{array}right).$$ The inverse should be clear from the definitions.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              I wrote down the details so I'm going to share just in case someone else needs it someday.



              The first thing to be noticed is that $sigma^{-1}({1})={1, p+2}$, hence:



              $$mathsf{Sh}(p+1, q)={sigmain mathsf{Sh}(p+1, q): sigma(1)=1}sqcup {sigmainmathsf{Sh}(p+1, q): sigma(p+2)=1}.$$
              This allows us to define the bijection easily setting:



              $$mathsf{Sh}(p, q)longrightarrow mathsf{Sh}(p+1, q)\ sigmalongmapsto left(begin{array}{ccccccc}1 & 2& ldots & p+2 & ldots &p+q+1\ 1 &sigma(1)+1 & ldots &sigma(p+1)+1 & ldots & sigma(p+q)+1 end{array}right)$$
              and



              $$ mathsf{Sh}(p+1, q-1)longrightarrow mathsf{Sh}(p+1, q)\ sigmalongmapsto left(begin{array}{ccccccc}1 & ldots &p+1 & p+2 & p+3& ldots & p+q+1\ sigma(1)+1 & ldots & sigma(p+1)+1 & 1 & sigma(p+2)+1 & ldots &sigma(p+q)+1 end{array}right).$$ The inverse should be clear from the definitions.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                I wrote down the details so I'm going to share just in case someone else needs it someday.



                The first thing to be noticed is that $sigma^{-1}({1})={1, p+2}$, hence:



                $$mathsf{Sh}(p+1, q)={sigmain mathsf{Sh}(p+1, q): sigma(1)=1}sqcup {sigmainmathsf{Sh}(p+1, q): sigma(p+2)=1}.$$
                This allows us to define the bijection easily setting:



                $$mathsf{Sh}(p, q)longrightarrow mathsf{Sh}(p+1, q)\ sigmalongmapsto left(begin{array}{ccccccc}1 & 2& ldots & p+2 & ldots &p+q+1\ 1 &sigma(1)+1 & ldots &sigma(p+1)+1 & ldots & sigma(p+q)+1 end{array}right)$$
                and



                $$ mathsf{Sh}(p+1, q-1)longrightarrow mathsf{Sh}(p+1, q)\ sigmalongmapsto left(begin{array}{ccccccc}1 & ldots &p+1 & p+2 & p+3& ldots & p+q+1\ sigma(1)+1 & ldots & sigma(p+1)+1 & 1 & sigma(p+2)+1 & ldots &sigma(p+q)+1 end{array}right).$$ The inverse should be clear from the definitions.






                share|cite|improve this answer









                $endgroup$



                I wrote down the details so I'm going to share just in case someone else needs it someday.



                The first thing to be noticed is that $sigma^{-1}({1})={1, p+2}$, hence:



                $$mathsf{Sh}(p+1, q)={sigmain mathsf{Sh}(p+1, q): sigma(1)=1}sqcup {sigmainmathsf{Sh}(p+1, q): sigma(p+2)=1}.$$
                This allows us to define the bijection easily setting:



                $$mathsf{Sh}(p, q)longrightarrow mathsf{Sh}(p+1, q)\ sigmalongmapsto left(begin{array}{ccccccc}1 & 2& ldots & p+2 & ldots &p+q+1\ 1 &sigma(1)+1 & ldots &sigma(p+1)+1 & ldots & sigma(p+q)+1 end{array}right)$$
                and



                $$ mathsf{Sh}(p+1, q-1)longrightarrow mathsf{Sh}(p+1, q)\ sigmalongmapsto left(begin{array}{ccccccc}1 & ldots &p+1 & p+2 & p+3& ldots & p+q+1\ sigma(1)+1 & ldots & sigma(p+1)+1 & 1 & sigma(p+2)+1 & ldots &sigma(p+q)+1 end{array}right).$$ The inverse should be clear from the definitions.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jun 8 '17 at 12:24









                PtFPtF

                3,94921733




                3,94921733






























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