Combinatorics Proof?











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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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  • I would say yes. Very much so.
    – Somos
    Dec 11 at 17:04










  • and in the last identity, you lost the Sum
    – G Cab
    Dec 11 at 17:45















up vote
3
down vote

favorite
2












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
























  • I would say yes. Very much so.
    – Somos
    Dec 11 at 17:04










  • and in the last identity, you lost the Sum
    – G Cab
    Dec 11 at 17:45













up vote
3
down vote

favorite
2









up vote
3
down vote

favorite
2






2





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















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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edited Dec 11 at 22:18

























asked Dec 11 at 16:52







user497933



















  • I would say yes. Very much so.
    – Somos
    Dec 11 at 17:04










  • and in the last identity, you lost the Sum
    – G Cab
    Dec 11 at 17:45


















  • I would say yes. Very much so.
    – Somos
    Dec 11 at 17:04










  • and in the last identity, you lost the Sum
    – G Cab
    Dec 11 at 17:45
















I would say yes. Very much so.
– Somos
Dec 11 at 17:04




I would say yes. Very much so.
– Somos
Dec 11 at 17:04












and in the last identity, you lost the Sum
– G Cab
Dec 11 at 17:45




and in the last identity, you lost the Sum
– G Cab
Dec 11 at 17:45










2 Answers
2






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6
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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}






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    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





















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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      active

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      active

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      up vote
      6
      down vote













      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

























        up vote
        6
        down vote













        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























          up vote
          6
          down vote










          up vote
          6
          down vote









          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












          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 at 17:21









          user9077

          1,234612




          1,234612






















              up vote
              2
              down vote













              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

























                up vote
                2
                down vote













                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























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  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












                  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 at 17:57









                  Mike Earnest

                  20k11950




                  20k11950






























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