Combinatorics Proof?












3












$begingroup$


How would I prove
$$sum_{b=0}^a frac{(2a)!}{b!b!(a-b)!(a-b)!} = binom{2a}{a}binom{2a}{a}$$



I am familiar with the identity $$binom{2n}{n}=sum_{k=0}^nbinom{n}{k}binom{n}{n-k}=binom{n}{k}^2$$ Would I be able to use this?






combinatorics proof-verification







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  • $begingroup$
    I would say yes. Very much so.
    $endgroup$
    – Somos
    Dec 11 '18 at 17:04










  • $begingroup$
    and in the last identity, you lost the Sum
    $endgroup$
    – G Cab
    Dec 11 '18 at 17:45
















3












$begingroup$


How would I prove
$$sum_{b=0}^a frac{(2a)!}{b!b!(a-b)!(a-b)!} = binom{2a}{a}binom{2a}{a}$$



I am familiar with the identity $$binom{2n}{n}=sum_{k=0}^nbinom{n}{k}binom{n}{n-k}=binom{n}{k}^2$$ Would I be able to use this?






combinatorics proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    I would say yes. Very much so.
    $endgroup$
    – Somos
    Dec 11 '18 at 17:04










  • $begingroup$
    and in the last identity, you lost the Sum
    $endgroup$
    – G Cab
    Dec 11 '18 at 17:45














3












3








3


2



$begingroup$


How would I prove
$$sum_{b=0}^a frac{(2a)!}{b!b!(a-b)!(a-b)!} = binom{2a}{a}binom{2a}{a}$$



I am familiar with the identity $$binom{2n}{n}=sum_{k=0}^nbinom{n}{k}binom{n}{n-k}=binom{n}{k}^2$$ Would I be able to use this?






combinatorics proof-verification







share|cite|improve this question











$endgroup$




How would I prove
$$sum_{b=0}^a frac{(2a)!}{b!b!(a-b)!(a-b)!} = binom{2a}{a}binom{2a}{a}$$



I am familiar with the identity $$binom{2n}{n}=sum_{k=0}^nbinom{n}{k}binom{n}{n-k}=binom{n}{k}^2$$ Would I be able to use this?






combinatorics proof-verification




combinatorics proof-verification






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 22:18

























asked Dec 11 '18 at 16:52







user497933



















  • $begingroup$
    I would say yes. Very much so.
    $endgroup$
    – Somos
    Dec 11 '18 at 17:04










  • $begingroup$
    and in the last identity, you lost the Sum
    $endgroup$
    – G Cab
    Dec 11 '18 at 17:45


















  • $begingroup$
    I would say yes. Very much so.
    $endgroup$
    – Somos
    Dec 11 '18 at 17:04










  • $begingroup$
    and in the last identity, you lost the Sum
    $endgroup$
    – G Cab
    Dec 11 '18 at 17:45
















$begingroup$
I would say yes. Very much so.
$endgroup$
– Somos
Dec 11 '18 at 17:04




$begingroup$
I would say yes. Very much so.
$endgroup$
– Somos
Dec 11 '18 at 17:04












$begingroup$
and in the last identity, you lost the Sum
$endgroup$
– G Cab
Dec 11 '18 at 17:45




$begingroup$
and in the last identity, you lost the Sum
$endgroup$
– G Cab
Dec 11 '18 at 17:45










2 Answers
2






active

oldest

votes


















6












$begingroup$

This is not purely bijective proof. The identity that you said you knew of it can be proved bijectively. We will use that fact to prove the new identity.



First rewrite
begin{align}
frac{(2a)!}{b!b!(a-b)!(a-b)!} &= frac{(2a)!}{a!a!}frac{a!a!}{b!(b-a)!b!(b-a)!}\
&={2a choose a}
{achoose b}{a choose b-a}end{align}
It follows that
begin{align}
sum_{b=0}^a frac{(2a)!}{b!b!(a-b)!(a-b)!} &={2a choose a}sum_{b=0}^a {achoose b}{achoose a-b}\
&={2achoose a}{2achoose a}quad text{(by your identity)}
end{align}






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    The summand $frac{(2a)!}{(b!)^2((a-b)!)^2}$ counts the number of ways to color the elements of a set $S$ of size $2a$ in four colors, so that there are $b$ elements in the first two colors and $a-b$ elements in the last two colors.



    The right hand side counts the number of ways to choose two subsets $A$ and $B$ of $S$ of size $a$. This induces a coloring in the following way:





    • $xin A, xin Bimplies x$ is red.


    • $xin A, xnotin Bimplies x$ is blue.


    • $xnotin A, xin Bimplies x$ is green.


    • $xnotin A, xnotin Bimplies x$ is black.


    Combining these two pieces, you can make a combinatorial proof.






    share|cite|improve this answer









    $endgroup$














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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      This is not purely bijective proof. The identity that you said you knew of it can be proved bijectively. We will use that fact to prove the new identity.



      First rewrite
      begin{align}
      frac{(2a)!}{b!b!(a-b)!(a-b)!} &= frac{(2a)!}{a!a!}frac{a!a!}{b!(b-a)!b!(b-a)!}\
      &={2a choose a}
      {achoose b}{a choose b-a}end{align}
      It follows that
      begin{align}
      sum_{b=0}^a frac{(2a)!}{b!b!(a-b)!(a-b)!} &={2a choose a}sum_{b=0}^a {achoose b}{achoose a-b}\
      &={2achoose a}{2achoose a}quad text{(by your identity)}
      end{align}






      share|cite|improve this answer









      $endgroup$


















        6












        $begingroup$

        This is not purely bijective proof. The identity that you said you knew of it can be proved bijectively. We will use that fact to prove the new identity.



        First rewrite
        begin{align}
        frac{(2a)!}{b!b!(a-b)!(a-b)!} &= frac{(2a)!}{a!a!}frac{a!a!}{b!(b-a)!b!(b-a)!}\
        &={2a choose a}
        {achoose b}{a choose b-a}end{align}
        It follows that
        begin{align}
        sum_{b=0}^a frac{(2a)!}{b!b!(a-b)!(a-b)!} &={2a choose a}sum_{b=0}^a {achoose b}{achoose a-b}\
        &={2achoose a}{2achoose a}quad text{(by your identity)}
        end{align}






        share|cite|improve this answer









        $endgroup$
















          6












          6








          6





          $begingroup$

          This is not purely bijective proof. The identity that you said you knew of it can be proved bijectively. We will use that fact to prove the new identity.



          First rewrite
          begin{align}
          frac{(2a)!}{b!b!(a-b)!(a-b)!} &= frac{(2a)!}{a!a!}frac{a!a!}{b!(b-a)!b!(b-a)!}\
          &={2a choose a}
          {achoose b}{a choose b-a}end{align}
          It follows that
          begin{align}
          sum_{b=0}^a frac{(2a)!}{b!b!(a-b)!(a-b)!} &={2a choose a}sum_{b=0}^a {achoose b}{achoose a-b}\
          &={2achoose a}{2achoose a}quad text{(by your identity)}
          end{align}






          share|cite|improve this answer









          $endgroup$



          This is not purely bijective proof. The identity that you said you knew of it can be proved bijectively. We will use that fact to prove the new identity.



          First rewrite
          begin{align}
          frac{(2a)!}{b!b!(a-b)!(a-b)!} &= frac{(2a)!}{a!a!}frac{a!a!}{b!(b-a)!b!(b-a)!}\
          &={2a choose a}
          {achoose b}{a choose b-a}end{align}
          It follows that
          begin{align}
          sum_{b=0}^a frac{(2a)!}{b!b!(a-b)!(a-b)!} &={2a choose a}sum_{b=0}^a {achoose b}{achoose a-b}\
          &={2achoose a}{2achoose a}quad text{(by your identity)}
          end{align}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 11 '18 at 17:21









          user9077user9077

          1,249612




          1,249612























              2












              $begingroup$

              The summand $frac{(2a)!}{(b!)^2((a-b)!)^2}$ counts the number of ways to color the elements of a set $S$ of size $2a$ in four colors, so that there are $b$ elements in the first two colors and $a-b$ elements in the last two colors.



              The right hand side counts the number of ways to choose two subsets $A$ and $B$ of $S$ of size $a$. This induces a coloring in the following way:





              • $xin A, xin Bimplies x$ is red.


              • $xin A, xnotin Bimplies x$ is blue.


              • $xnotin A, xin Bimplies x$ is green.


              • $xnotin A, xnotin Bimplies x$ is black.


              Combining these two pieces, you can make a combinatorial proof.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                The summand $frac{(2a)!}{(b!)^2((a-b)!)^2}$ counts the number of ways to color the elements of a set $S$ of size $2a$ in four colors, so that there are $b$ elements in the first two colors and $a-b$ elements in the last two colors.



                The right hand side counts the number of ways to choose two subsets $A$ and $B$ of $S$ of size $a$. This induces a coloring in the following way:





                • $xin A, xin Bimplies x$ is red.


                • $xin A, xnotin Bimplies x$ is blue.


                • $xnotin A, xin Bimplies x$ is green.


                • $xnotin A, xnotin Bimplies x$ is black.


                Combining these two pieces, you can make a combinatorial proof.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  The summand $frac{(2a)!}{(b!)^2((a-b)!)^2}$ counts the number of ways to color the elements of a set $S$ of size $2a$ in four colors, so that there are $b$ elements in the first two colors and $a-b$ elements in the last two colors.



                  The right hand side counts the number of ways to choose two subsets $A$ and $B$ of $S$ of size $a$. This induces a coloring in the following way:





                  • $xin A, xin Bimplies x$ is red.


                  • $xin A, xnotin Bimplies x$ is blue.


                  • $xnotin A, xin Bimplies x$ is green.


                  • $xnotin A, xnotin Bimplies x$ is black.


                  Combining these two pieces, you can make a combinatorial proof.






                  share|cite|improve this answer









                  $endgroup$



                  The summand $frac{(2a)!}{(b!)^2((a-b)!)^2}$ counts the number of ways to color the elements of a set $S$ of size $2a$ in four colors, so that there are $b$ elements in the first two colors and $a-b$ elements in the last two colors.



                  The right hand side counts the number of ways to choose two subsets $A$ and $B$ of $S$ of size $a$. This induces a coloring in the following way:





                  • $xin A, xin Bimplies x$ is red.


                  • $xin A, xnotin Bimplies x$ is blue.


                  • $xnotin A, xin Bimplies x$ is green.


                  • $xnotin A, xnotin Bimplies x$ is black.


                  Combining these two pieces, you can make a combinatorial proof.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 11 '18 at 17:57









                  Mike EarnestMike Earnest

                  26.5k22151




                  26.5k22151






























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