Combinatorial interpretation of identity: $sumlimits_{j=0}^bbinom{b}{j}^2binom{n+j}{2b}=binom{n}{b}^2$











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Currently, I am trying to prove the following two identities, which arose as a result of my other question in the Math StackExchange recently:



begin{equation}
sum_{j=0}^bbinom{b}{j}^2binom{n+j}{2b}=binom{n}{b}^2
end{equation}



begin{equation}
sum_{j=0}^b(-1)^jbinom{b}{j}binom{n+j}{2b} = left(-1right)^b binom{n}{b}
end{equation}



In particular, the first identity appeared as a remark in Volume 4 of Henry Gould's Combinatorial Identities.



Out of curiosity, I am wondering if there is a combinatorial proof for any of these two identities?



EDIT: I have managed to prove the second identity, by considering the coefficient of $x^{n-b}$ in the expansion of $(1+x)^{-(b+1)}=(1+x)^b(1+x)^{-(2b+1)}$.



EDIT 2: I have also managed to find a proof for the first identity by using the Vandermonde identity. I would still be interested however, in a proof of the second identity using a more combinatorial approach.










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  • I'm not sure if this is going anywhere at all, but in the first identity, the RHS is the number of ways to pick two subsets of ${ 1,2, ldots , n }$ with $b$ elements. I was trying to come up with a bijection to get to the LHS; my intuition is that maybe the $j$ on the LHS represents the number of elements that the two sets have in common, but I wasn't able to get anywhere with that.
    – Marcus M
    Apr 15 '15 at 2:48















up vote
14
down vote

favorite
9












Currently, I am trying to prove the following two identities, which arose as a result of my other question in the Math StackExchange recently:



begin{equation}
sum_{j=0}^bbinom{b}{j}^2binom{n+j}{2b}=binom{n}{b}^2
end{equation}



begin{equation}
sum_{j=0}^b(-1)^jbinom{b}{j}binom{n+j}{2b} = left(-1right)^b binom{n}{b}
end{equation}



In particular, the first identity appeared as a remark in Volume 4 of Henry Gould's Combinatorial Identities.



Out of curiosity, I am wondering if there is a combinatorial proof for any of these two identities?



EDIT: I have managed to prove the second identity, by considering the coefficient of $x^{n-b}$ in the expansion of $(1+x)^{-(b+1)}=(1+x)^b(1+x)^{-(2b+1)}$.



EDIT 2: I have also managed to find a proof for the first identity by using the Vandermonde identity. I would still be interested however, in a proof of the second identity using a more combinatorial approach.










share|cite|improve this question
























  • I'm not sure if this is going anywhere at all, but in the first identity, the RHS is the number of ways to pick two subsets of ${ 1,2, ldots , n }$ with $b$ elements. I was trying to come up with a bijection to get to the LHS; my intuition is that maybe the $j$ on the LHS represents the number of elements that the two sets have in common, but I wasn't able to get anywhere with that.
    – Marcus M
    Apr 15 '15 at 2:48













up vote
14
down vote

favorite
9









up vote
14
down vote

favorite
9






9





Currently, I am trying to prove the following two identities, which arose as a result of my other question in the Math StackExchange recently:



begin{equation}
sum_{j=0}^bbinom{b}{j}^2binom{n+j}{2b}=binom{n}{b}^2
end{equation}



begin{equation}
sum_{j=0}^b(-1)^jbinom{b}{j}binom{n+j}{2b} = left(-1right)^b binom{n}{b}
end{equation}



In particular, the first identity appeared as a remark in Volume 4 of Henry Gould's Combinatorial Identities.



Out of curiosity, I am wondering if there is a combinatorial proof for any of these two identities?



EDIT: I have managed to prove the second identity, by considering the coefficient of $x^{n-b}$ in the expansion of $(1+x)^{-(b+1)}=(1+x)^b(1+x)^{-(2b+1)}$.



EDIT 2: I have also managed to find a proof for the first identity by using the Vandermonde identity. I would still be interested however, in a proof of the second identity using a more combinatorial approach.










share|cite|improve this question















Currently, I am trying to prove the following two identities, which arose as a result of my other question in the Math StackExchange recently:



begin{equation}
sum_{j=0}^bbinom{b}{j}^2binom{n+j}{2b}=binom{n}{b}^2
end{equation}



begin{equation}
sum_{j=0}^b(-1)^jbinom{b}{j}binom{n+j}{2b} = left(-1right)^b binom{n}{b}
end{equation}



In particular, the first identity appeared as a remark in Volume 4 of Henry Gould's Combinatorial Identities.



Out of curiosity, I am wondering if there is a combinatorial proof for any of these two identities?



EDIT: I have managed to prove the second identity, by considering the coefficient of $x^{n-b}$ in the expansion of $(1+x)^{-(b+1)}=(1+x)^b(1+x)^{-(2b+1)}$.



EDIT 2: I have also managed to find a proof for the first identity by using the Vandermonde identity. I would still be interested however, in a proof of the second identity using a more combinatorial approach.







combinatorics binomial-coefficients combinatorial-proofs






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edited Nov 23 at 18:36









darij grinberg

10.1k32961




10.1k32961










asked Apr 14 '15 at 10:44









yoshi

426212




426212












  • I'm not sure if this is going anywhere at all, but in the first identity, the RHS is the number of ways to pick two subsets of ${ 1,2, ldots , n }$ with $b$ elements. I was trying to come up with a bijection to get to the LHS; my intuition is that maybe the $j$ on the LHS represents the number of elements that the two sets have in common, but I wasn't able to get anywhere with that.
    – Marcus M
    Apr 15 '15 at 2:48


















  • I'm not sure if this is going anywhere at all, but in the first identity, the RHS is the number of ways to pick two subsets of ${ 1,2, ldots , n }$ with $b$ elements. I was trying to come up with a bijection to get to the LHS; my intuition is that maybe the $j$ on the LHS represents the number of elements that the two sets have in common, but I wasn't able to get anywhere with that.
    – Marcus M
    Apr 15 '15 at 2:48
















I'm not sure if this is going anywhere at all, but in the first identity, the RHS is the number of ways to pick two subsets of ${ 1,2, ldots , n }$ with $b$ elements. I was trying to come up with a bijection to get to the LHS; my intuition is that maybe the $j$ on the LHS represents the number of elements that the two sets have in common, but I wasn't able to get anywhere with that.
– Marcus M
Apr 15 '15 at 2:48




I'm not sure if this is going anywhere at all, but in the first identity, the RHS is the number of ways to pick two subsets of ${ 1,2, ldots , n }$ with $b$ elements. I was trying to come up with a bijection to get to the LHS; my intuition is that maybe the $j$ on the LHS represents the number of elements that the two sets have in common, but I wasn't able to get anywhere with that.
– Marcus M
Apr 15 '15 at 2:48










3 Answers
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3
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The first identity is the particular case (for $a=b$ and $x=n$) of the identity
$$
sum_{i=0}^b dbinom{a}{i} dbinom{b}{i} dbinom{x+i}{a+b} = dbinom{x}{a} dbinom{x}{b} ,
$$

which has appeared in math.stackexchange question #280481. For combinatorial proofs, see




  • George E. Andrews, Identities in combinatorics, I: on sorting two ordered sets, Discrete Mathematics, Volume 11, Issue 2, 1975, Pages 97--106.


  • George E. Andrews, David M. Bressoud, Identities in combinatorics III: Further aspects of ordered set sorting, Discrete Mathematics, Volume 49, Issue 3, May 1984, Pages 223--236.



I also give an algebraic proof in Proposition 3.39 (f) of my Notes on the combinatorial fundamentals of algebra (search for "assortment of other identities" if the numbering has changed, or else look at the archived version of 7 November 2018).





The second identity is the particular case (for $d=2b$) of the following identity:
$$
sum_{j=0}^b left(-1right)^j dbinom{b}{j} dbinom{n+j}{d} = left(-1right)^b dbinom{n}{d-b} ,
$$

where $n$ and $b$ are two nonnegative integers and $d$ is an integer.
This is some applications of symmetry and upper negation away from the Vandermonde convolution (and can also be easily proven by induction over $b$). This does not give a combinatorial proof, but it makes me suspect that it is well-known, and hopefully someone has collected combinatorial proofs of such identities.






share|cite|improve this answer























  • (+1). The references and observations are a good improvement of the page.
    – Marko Riedel
    Sep 28 '15 at 1:44


















up vote
2
down vote













By way of enrichment permit me to present an algebraic proof.



Suppose we seek to verify that
$$sum_{j=0}^b {bchoose j}^2
{n+jchoose 2b} = {nchoose b}^2.$$
where $0le ble n.$



Introduce
$${bchoose j} =
frac{1}{2pi i}
int_{|z|=epsilon} frac{1}{z^{b-j+1}}
frac{1}{(1-z)^{j+1}} ; dz$$



and
$${n+jchoose 2b} =
frac{1}{2pi i}
int_{|w|=epsilon} frac{1}{w^{2b+1}}
(1+w)^{n+j} ; dw.$$



This yields for the sum
$$frac{1}{2pi i}
int_{|w|=epsilon} frac{(1+w)^{n}}{w^{2b+1}}
frac{1}{2pi i}
int_{|z|=epsilon} frac{1}{z^{b+1}}
frac{1}{1-z}
sum_{j=0}^b {bchoose j} (1+w)^j frac{z^j}{(1-z)^j}
; dz ; dw
\ = frac{1}{2pi i}
int_{|w|=epsilon} frac{(1+w)^{n}}{w^{2b+1}}
frac{1}{2pi i}
int_{|z|=epsilon} frac{1}{z^{b+1}}
frac{1}{1-z}
left(1+frac{(1+w)z}{1-z}right)^b
; dz ; dw
\ = frac{1}{2pi i}
int_{|w|=epsilon} frac{(1+w)^{n}}{w^{2b+1}}
frac{1}{2pi i}
int_{|z|=epsilon} frac{1}{z^{b+1}}
frac{1}{(1-z)^{b+1}}
(1+wz)^b ; dz ; dw.$$



The inner residue is
$$sum_{q=0}^b {bchoose q} w^q {b-q+bchoose b}.$$
Substitute this into the outer integral to get
$$sum_{q=0}^b {bchoose q} {2b-qchoose b}
{nchoose 2b-q}.$$



Observe that
$${2b-qchoose b}{nchoose 2b-q}
= frac{(2b-q)!}{b! (b-q)!}frac{n!}{(2b-q)! (n-2b+q)!}
\ = frac{(n-b)!}{b! (b-q)!}frac{n!}{(n-b)! (n-2b+q)!}
= {nchoose b} {n-bchoose b-q}.$$



This yields for the sum
$${nchoose b} sum_{q=0}^b {bchoose q}
{n-bchoose b-q}.$$



which evaluates to
$${nchoose b}^2$$
by inspection.



It can also be done with the integral
$$frac{1}{2pi i}
int_{|z|=epsilon} frac{1}{z^{b-q+1}} (1+z)^{n-b} ; dz$$



which yields
$${nchoose b} frac{1}{2pi i}
int_{|z|=epsilon} frac{1}{z^{b+1}} (1+z)^{n-b}
sum_{q=0}^b {bchoose q} z^q ; dz
\ = {nchoose b} frac{1}{2pi i}
int_{|z|=epsilon} frac{1}{z^{b+1}} (1+z)^{n} ; dz
= {nchoose b}^2.$$






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    Combinatorial proof of the second identity



    Let $B$ be a $b$-element set and let $N$ be an $n$-element set disjoint to $B$.



    The sum $sum_jbinom b{b-j}binom{n+j}{2b}$ counts the number of pairs of sets $Tsubset B$, $Xsubset Ncup B$ s.t. $|X|=2b$ and $Xcap T=varnothing$. (Indeed, one summand corresponds to choosing first $(b-j)$-element set $Tsubset B$ and then $2b$-element set $Xsubset Ncup(Bsetminus T)$.) And LHS counts such pairs with signs $(-1)^{|B setminus T|}$.



    Now there is a sign-reversing involution: find the smallest element of $Bsetminus X$ and then add it to $T$ if it's not already in $T$, otherwise remove it from $T$. So in LHS everything with non-empty $Bsetminus X$ cancels out; and there are exactly $binom nb$ variants with $Xsupset B$ and they are counted with sign $(-1)^b$.



    (And the RHS is, in fact, $(-1)^bbinom nb$.)






    share|cite|improve this answer



















    • 1




      I guess this argument also applies to the generalization $sumlimits_{j=0}^b left(-1right)^j dbinom{b}{j} dbinom{n+j}{d} = left(-1right)^b dbinom{n}{d-b}$ ?
      – darij grinberg
      Sep 28 '15 at 1:48










    • (Re: generalization) indeed
      – Grigory M
      Oct 1 '15 at 14:14











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    3 Answers
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    3 Answers
    3






    active

    oldest

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    active

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    active

    oldest

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    up vote
    3
    down vote













    The first identity is the particular case (for $a=b$ and $x=n$) of the identity
    $$
    sum_{i=0}^b dbinom{a}{i} dbinom{b}{i} dbinom{x+i}{a+b} = dbinom{x}{a} dbinom{x}{b} ,
    $$

    which has appeared in math.stackexchange question #280481. For combinatorial proofs, see




    • George E. Andrews, Identities in combinatorics, I: on sorting two ordered sets, Discrete Mathematics, Volume 11, Issue 2, 1975, Pages 97--106.


    • George E. Andrews, David M. Bressoud, Identities in combinatorics III: Further aspects of ordered set sorting, Discrete Mathematics, Volume 49, Issue 3, May 1984, Pages 223--236.



    I also give an algebraic proof in Proposition 3.39 (f) of my Notes on the combinatorial fundamentals of algebra (search for "assortment of other identities" if the numbering has changed, or else look at the archived version of 7 November 2018).





    The second identity is the particular case (for $d=2b$) of the following identity:
    $$
    sum_{j=0}^b left(-1right)^j dbinom{b}{j} dbinom{n+j}{d} = left(-1right)^b dbinom{n}{d-b} ,
    $$

    where $n$ and $b$ are two nonnegative integers and $d$ is an integer.
    This is some applications of symmetry and upper negation away from the Vandermonde convolution (and can also be easily proven by induction over $b$). This does not give a combinatorial proof, but it makes me suspect that it is well-known, and hopefully someone has collected combinatorial proofs of such identities.






    share|cite|improve this answer























    • (+1). The references and observations are a good improvement of the page.
      – Marko Riedel
      Sep 28 '15 at 1:44















    up vote
    3
    down vote













    The first identity is the particular case (for $a=b$ and $x=n$) of the identity
    $$
    sum_{i=0}^b dbinom{a}{i} dbinom{b}{i} dbinom{x+i}{a+b} = dbinom{x}{a} dbinom{x}{b} ,
    $$

    which has appeared in math.stackexchange question #280481. For combinatorial proofs, see




    • George E. Andrews, Identities in combinatorics, I: on sorting two ordered sets, Discrete Mathematics, Volume 11, Issue 2, 1975, Pages 97--106.


    • George E. Andrews, David M. Bressoud, Identities in combinatorics III: Further aspects of ordered set sorting, Discrete Mathematics, Volume 49, Issue 3, May 1984, Pages 223--236.



    I also give an algebraic proof in Proposition 3.39 (f) of my Notes on the combinatorial fundamentals of algebra (search for "assortment of other identities" if the numbering has changed, or else look at the archived version of 7 November 2018).





    The second identity is the particular case (for $d=2b$) of the following identity:
    $$
    sum_{j=0}^b left(-1right)^j dbinom{b}{j} dbinom{n+j}{d} = left(-1right)^b dbinom{n}{d-b} ,
    $$

    where $n$ and $b$ are two nonnegative integers and $d$ is an integer.
    This is some applications of symmetry and upper negation away from the Vandermonde convolution (and can also be easily proven by induction over $b$). This does not give a combinatorial proof, but it makes me suspect that it is well-known, and hopefully someone has collected combinatorial proofs of such identities.






    share|cite|improve this answer























    • (+1). The references and observations are a good improvement of the page.
      – Marko Riedel
      Sep 28 '15 at 1:44













    up vote
    3
    down vote










    up vote
    3
    down vote









    The first identity is the particular case (for $a=b$ and $x=n$) of the identity
    $$
    sum_{i=0}^b dbinom{a}{i} dbinom{b}{i} dbinom{x+i}{a+b} = dbinom{x}{a} dbinom{x}{b} ,
    $$

    which has appeared in math.stackexchange question #280481. For combinatorial proofs, see




    • George E. Andrews, Identities in combinatorics, I: on sorting two ordered sets, Discrete Mathematics, Volume 11, Issue 2, 1975, Pages 97--106.


    • George E. Andrews, David M. Bressoud, Identities in combinatorics III: Further aspects of ordered set sorting, Discrete Mathematics, Volume 49, Issue 3, May 1984, Pages 223--236.



    I also give an algebraic proof in Proposition 3.39 (f) of my Notes on the combinatorial fundamentals of algebra (search for "assortment of other identities" if the numbering has changed, or else look at the archived version of 7 November 2018).





    The second identity is the particular case (for $d=2b$) of the following identity:
    $$
    sum_{j=0}^b left(-1right)^j dbinom{b}{j} dbinom{n+j}{d} = left(-1right)^b dbinom{n}{d-b} ,
    $$

    where $n$ and $b$ are two nonnegative integers and $d$ is an integer.
    This is some applications of symmetry and upper negation away from the Vandermonde convolution (and can also be easily proven by induction over $b$). This does not give a combinatorial proof, but it makes me suspect that it is well-known, and hopefully someone has collected combinatorial proofs of such identities.






    share|cite|improve this answer














    The first identity is the particular case (for $a=b$ and $x=n$) of the identity
    $$
    sum_{i=0}^b dbinom{a}{i} dbinom{b}{i} dbinom{x+i}{a+b} = dbinom{x}{a} dbinom{x}{b} ,
    $$

    which has appeared in math.stackexchange question #280481. For combinatorial proofs, see




    • George E. Andrews, Identities in combinatorics, I: on sorting two ordered sets, Discrete Mathematics, Volume 11, Issue 2, 1975, Pages 97--106.


    • George E. Andrews, David M. Bressoud, Identities in combinatorics III: Further aspects of ordered set sorting, Discrete Mathematics, Volume 49, Issue 3, May 1984, Pages 223--236.



    I also give an algebraic proof in Proposition 3.39 (f) of my Notes on the combinatorial fundamentals of algebra (search for "assortment of other identities" if the numbering has changed, or else look at the archived version of 7 November 2018).





    The second identity is the particular case (for $d=2b$) of the following identity:
    $$
    sum_{j=0}^b left(-1right)^j dbinom{b}{j} dbinom{n+j}{d} = left(-1right)^b dbinom{n}{d-b} ,
    $$

    where $n$ and $b$ are two nonnegative integers and $d$ is an integer.
    This is some applications of symmetry and upper negation away from the Vandermonde convolution (and can also be easily proven by induction over $b$). This does not give a combinatorial proof, but it makes me suspect that it is well-known, and hopefully someone has collected combinatorial proofs of such identities.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 23 at 18:35

























    answered Sep 28 '15 at 1:21









    darij grinberg

    10.1k32961




    10.1k32961












    • (+1). The references and observations are a good improvement of the page.
      – Marko Riedel
      Sep 28 '15 at 1:44


















    • (+1). The references and observations are a good improvement of the page.
      – Marko Riedel
      Sep 28 '15 at 1:44
















    (+1). The references and observations are a good improvement of the page.
    – Marko Riedel
    Sep 28 '15 at 1:44




    (+1). The references and observations are a good improvement of the page.
    – Marko Riedel
    Sep 28 '15 at 1:44










    up vote
    2
    down vote













    By way of enrichment permit me to present an algebraic proof.



    Suppose we seek to verify that
    $$sum_{j=0}^b {bchoose j}^2
    {n+jchoose 2b} = {nchoose b}^2.$$
    where $0le ble n.$



    Introduce
    $${bchoose j} =
    frac{1}{2pi i}
    int_{|z|=epsilon} frac{1}{z^{b-j+1}}
    frac{1}{(1-z)^{j+1}} ; dz$$



    and
    $${n+jchoose 2b} =
    frac{1}{2pi i}
    int_{|w|=epsilon} frac{1}{w^{2b+1}}
    (1+w)^{n+j} ; dw.$$



    This yields for the sum
    $$frac{1}{2pi i}
    int_{|w|=epsilon} frac{(1+w)^{n}}{w^{2b+1}}
    frac{1}{2pi i}
    int_{|z|=epsilon} frac{1}{z^{b+1}}
    frac{1}{1-z}
    sum_{j=0}^b {bchoose j} (1+w)^j frac{z^j}{(1-z)^j}
    ; dz ; dw
    \ = frac{1}{2pi i}
    int_{|w|=epsilon} frac{(1+w)^{n}}{w^{2b+1}}
    frac{1}{2pi i}
    int_{|z|=epsilon} frac{1}{z^{b+1}}
    frac{1}{1-z}
    left(1+frac{(1+w)z}{1-z}right)^b
    ; dz ; dw
    \ = frac{1}{2pi i}
    int_{|w|=epsilon} frac{(1+w)^{n}}{w^{2b+1}}
    frac{1}{2pi i}
    int_{|z|=epsilon} frac{1}{z^{b+1}}
    frac{1}{(1-z)^{b+1}}
    (1+wz)^b ; dz ; dw.$$



    The inner residue is
    $$sum_{q=0}^b {bchoose q} w^q {b-q+bchoose b}.$$
    Substitute this into the outer integral to get
    $$sum_{q=0}^b {bchoose q} {2b-qchoose b}
    {nchoose 2b-q}.$$



    Observe that
    $${2b-qchoose b}{nchoose 2b-q}
    = frac{(2b-q)!}{b! (b-q)!}frac{n!}{(2b-q)! (n-2b+q)!}
    \ = frac{(n-b)!}{b! (b-q)!}frac{n!}{(n-b)! (n-2b+q)!}
    = {nchoose b} {n-bchoose b-q}.$$



    This yields for the sum
    $${nchoose b} sum_{q=0}^b {bchoose q}
    {n-bchoose b-q}.$$



    which evaluates to
    $${nchoose b}^2$$
    by inspection.



    It can also be done with the integral
    $$frac{1}{2pi i}
    int_{|z|=epsilon} frac{1}{z^{b-q+1}} (1+z)^{n-b} ; dz$$



    which yields
    $${nchoose b} frac{1}{2pi i}
    int_{|z|=epsilon} frac{1}{z^{b+1}} (1+z)^{n-b}
    sum_{q=0}^b {bchoose q} z^q ; dz
    \ = {nchoose b} frac{1}{2pi i}
    int_{|z|=epsilon} frac{1}{z^{b+1}} (1+z)^{n} ; dz
    = {nchoose b}^2.$$






    share|cite|improve this answer



























      up vote
      2
      down vote













      By way of enrichment permit me to present an algebraic proof.



      Suppose we seek to verify that
      $$sum_{j=0}^b {bchoose j}^2
      {n+jchoose 2b} = {nchoose b}^2.$$
      where $0le ble n.$



      Introduce
      $${bchoose j} =
      frac{1}{2pi i}
      int_{|z|=epsilon} frac{1}{z^{b-j+1}}
      frac{1}{(1-z)^{j+1}} ; dz$$



      and
      $${n+jchoose 2b} =
      frac{1}{2pi i}
      int_{|w|=epsilon} frac{1}{w^{2b+1}}
      (1+w)^{n+j} ; dw.$$



      This yields for the sum
      $$frac{1}{2pi i}
      int_{|w|=epsilon} frac{(1+w)^{n}}{w^{2b+1}}
      frac{1}{2pi i}
      int_{|z|=epsilon} frac{1}{z^{b+1}}
      frac{1}{1-z}
      sum_{j=0}^b {bchoose j} (1+w)^j frac{z^j}{(1-z)^j}
      ; dz ; dw
      \ = frac{1}{2pi i}
      int_{|w|=epsilon} frac{(1+w)^{n}}{w^{2b+1}}
      frac{1}{2pi i}
      int_{|z|=epsilon} frac{1}{z^{b+1}}
      frac{1}{1-z}
      left(1+frac{(1+w)z}{1-z}right)^b
      ; dz ; dw
      \ = frac{1}{2pi i}
      int_{|w|=epsilon} frac{(1+w)^{n}}{w^{2b+1}}
      frac{1}{2pi i}
      int_{|z|=epsilon} frac{1}{z^{b+1}}
      frac{1}{(1-z)^{b+1}}
      (1+wz)^b ; dz ; dw.$$



      The inner residue is
      $$sum_{q=0}^b {bchoose q} w^q {b-q+bchoose b}.$$
      Substitute this into the outer integral to get
      $$sum_{q=0}^b {bchoose q} {2b-qchoose b}
      {nchoose 2b-q}.$$



      Observe that
      $${2b-qchoose b}{nchoose 2b-q}
      = frac{(2b-q)!}{b! (b-q)!}frac{n!}{(2b-q)! (n-2b+q)!}
      \ = frac{(n-b)!}{b! (b-q)!}frac{n!}{(n-b)! (n-2b+q)!}
      = {nchoose b} {n-bchoose b-q}.$$



      This yields for the sum
      $${nchoose b} sum_{q=0}^b {bchoose q}
      {n-bchoose b-q}.$$



      which evaluates to
      $${nchoose b}^2$$
      by inspection.



      It can also be done with the integral
      $$frac{1}{2pi i}
      int_{|z|=epsilon} frac{1}{z^{b-q+1}} (1+z)^{n-b} ; dz$$



      which yields
      $${nchoose b} frac{1}{2pi i}
      int_{|z|=epsilon} frac{1}{z^{b+1}} (1+z)^{n-b}
      sum_{q=0}^b {bchoose q} z^q ; dz
      \ = {nchoose b} frac{1}{2pi i}
      int_{|z|=epsilon} frac{1}{z^{b+1}} (1+z)^{n} ; dz
      = {nchoose b}^2.$$






      share|cite|improve this answer

























        up vote
        2
        down vote










        up vote
        2
        down vote









        By way of enrichment permit me to present an algebraic proof.



        Suppose we seek to verify that
        $$sum_{j=0}^b {bchoose j}^2
        {n+jchoose 2b} = {nchoose b}^2.$$
        where $0le ble n.$



        Introduce
        $${bchoose j} =
        frac{1}{2pi i}
        int_{|z|=epsilon} frac{1}{z^{b-j+1}}
        frac{1}{(1-z)^{j+1}} ; dz$$



        and
        $${n+jchoose 2b} =
        frac{1}{2pi i}
        int_{|w|=epsilon} frac{1}{w^{2b+1}}
        (1+w)^{n+j} ; dw.$$



        This yields for the sum
        $$frac{1}{2pi i}
        int_{|w|=epsilon} frac{(1+w)^{n}}{w^{2b+1}}
        frac{1}{2pi i}
        int_{|z|=epsilon} frac{1}{z^{b+1}}
        frac{1}{1-z}
        sum_{j=0}^b {bchoose j} (1+w)^j frac{z^j}{(1-z)^j}
        ; dz ; dw
        \ = frac{1}{2pi i}
        int_{|w|=epsilon} frac{(1+w)^{n}}{w^{2b+1}}
        frac{1}{2pi i}
        int_{|z|=epsilon} frac{1}{z^{b+1}}
        frac{1}{1-z}
        left(1+frac{(1+w)z}{1-z}right)^b
        ; dz ; dw
        \ = frac{1}{2pi i}
        int_{|w|=epsilon} frac{(1+w)^{n}}{w^{2b+1}}
        frac{1}{2pi i}
        int_{|z|=epsilon} frac{1}{z^{b+1}}
        frac{1}{(1-z)^{b+1}}
        (1+wz)^b ; dz ; dw.$$



        The inner residue is
        $$sum_{q=0}^b {bchoose q} w^q {b-q+bchoose b}.$$
        Substitute this into the outer integral to get
        $$sum_{q=0}^b {bchoose q} {2b-qchoose b}
        {nchoose 2b-q}.$$



        Observe that
        $${2b-qchoose b}{nchoose 2b-q}
        = frac{(2b-q)!}{b! (b-q)!}frac{n!}{(2b-q)! (n-2b+q)!}
        \ = frac{(n-b)!}{b! (b-q)!}frac{n!}{(n-b)! (n-2b+q)!}
        = {nchoose b} {n-bchoose b-q}.$$



        This yields for the sum
        $${nchoose b} sum_{q=0}^b {bchoose q}
        {n-bchoose b-q}.$$



        which evaluates to
        $${nchoose b}^2$$
        by inspection.



        It can also be done with the integral
        $$frac{1}{2pi i}
        int_{|z|=epsilon} frac{1}{z^{b-q+1}} (1+z)^{n-b} ; dz$$



        which yields
        $${nchoose b} frac{1}{2pi i}
        int_{|z|=epsilon} frac{1}{z^{b+1}} (1+z)^{n-b}
        sum_{q=0}^b {bchoose q} z^q ; dz
        \ = {nchoose b} frac{1}{2pi i}
        int_{|z|=epsilon} frac{1}{z^{b+1}} (1+z)^{n} ; dz
        = {nchoose b}^2.$$






        share|cite|improve this answer














        By way of enrichment permit me to present an algebraic proof.



        Suppose we seek to verify that
        $$sum_{j=0}^b {bchoose j}^2
        {n+jchoose 2b} = {nchoose b}^2.$$
        where $0le ble n.$



        Introduce
        $${bchoose j} =
        frac{1}{2pi i}
        int_{|z|=epsilon} frac{1}{z^{b-j+1}}
        frac{1}{(1-z)^{j+1}} ; dz$$



        and
        $${n+jchoose 2b} =
        frac{1}{2pi i}
        int_{|w|=epsilon} frac{1}{w^{2b+1}}
        (1+w)^{n+j} ; dw.$$



        This yields for the sum
        $$frac{1}{2pi i}
        int_{|w|=epsilon} frac{(1+w)^{n}}{w^{2b+1}}
        frac{1}{2pi i}
        int_{|z|=epsilon} frac{1}{z^{b+1}}
        frac{1}{1-z}
        sum_{j=0}^b {bchoose j} (1+w)^j frac{z^j}{(1-z)^j}
        ; dz ; dw
        \ = frac{1}{2pi i}
        int_{|w|=epsilon} frac{(1+w)^{n}}{w^{2b+1}}
        frac{1}{2pi i}
        int_{|z|=epsilon} frac{1}{z^{b+1}}
        frac{1}{1-z}
        left(1+frac{(1+w)z}{1-z}right)^b
        ; dz ; dw
        \ = frac{1}{2pi i}
        int_{|w|=epsilon} frac{(1+w)^{n}}{w^{2b+1}}
        frac{1}{2pi i}
        int_{|z|=epsilon} frac{1}{z^{b+1}}
        frac{1}{(1-z)^{b+1}}
        (1+wz)^b ; dz ; dw.$$



        The inner residue is
        $$sum_{q=0}^b {bchoose q} w^q {b-q+bchoose b}.$$
        Substitute this into the outer integral to get
        $$sum_{q=0}^b {bchoose q} {2b-qchoose b}
        {nchoose 2b-q}.$$



        Observe that
        $${2b-qchoose b}{nchoose 2b-q}
        = frac{(2b-q)!}{b! (b-q)!}frac{n!}{(2b-q)! (n-2b+q)!}
        \ = frac{(n-b)!}{b! (b-q)!}frac{n!}{(n-b)! (n-2b+q)!}
        = {nchoose b} {n-bchoose b-q}.$$



        This yields for the sum
        $${nchoose b} sum_{q=0}^b {bchoose q}
        {n-bchoose b-q}.$$



        which evaluates to
        $${nchoose b}^2$$
        by inspection.



        It can also be done with the integral
        $$frac{1}{2pi i}
        int_{|z|=epsilon} frac{1}{z^{b-q+1}} (1+z)^{n-b} ; dz$$



        which yields
        $${nchoose b} frac{1}{2pi i}
        int_{|z|=epsilon} frac{1}{z^{b+1}} (1+z)^{n-b}
        sum_{q=0}^b {bchoose q} z^q ; dz
        \ = {nchoose b} frac{1}{2pi i}
        int_{|z|=epsilon} frac{1}{z^{b+1}} (1+z)^{n} ; dz
        = {nchoose b}^2.$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Sep 28 '15 at 0:24

























        answered Sep 27 '15 at 23:44









        Marko Riedel

        38.5k339106




        38.5k339106






















            up vote
            2
            down vote













            Combinatorial proof of the second identity



            Let $B$ be a $b$-element set and let $N$ be an $n$-element set disjoint to $B$.



            The sum $sum_jbinom b{b-j}binom{n+j}{2b}$ counts the number of pairs of sets $Tsubset B$, $Xsubset Ncup B$ s.t. $|X|=2b$ and $Xcap T=varnothing$. (Indeed, one summand corresponds to choosing first $(b-j)$-element set $Tsubset B$ and then $2b$-element set $Xsubset Ncup(Bsetminus T)$.) And LHS counts such pairs with signs $(-1)^{|B setminus T|}$.



            Now there is a sign-reversing involution: find the smallest element of $Bsetminus X$ and then add it to $T$ if it's not already in $T$, otherwise remove it from $T$. So in LHS everything with non-empty $Bsetminus X$ cancels out; and there are exactly $binom nb$ variants with $Xsupset B$ and they are counted with sign $(-1)^b$.



            (And the RHS is, in fact, $(-1)^bbinom nb$.)






            share|cite|improve this answer



















            • 1




              I guess this argument also applies to the generalization $sumlimits_{j=0}^b left(-1right)^j dbinom{b}{j} dbinom{n+j}{d} = left(-1right)^b dbinom{n}{d-b}$ ?
              – darij grinberg
              Sep 28 '15 at 1:48










            • (Re: generalization) indeed
              – Grigory M
              Oct 1 '15 at 14:14















            up vote
            2
            down vote













            Combinatorial proof of the second identity



            Let $B$ be a $b$-element set and let $N$ be an $n$-element set disjoint to $B$.



            The sum $sum_jbinom b{b-j}binom{n+j}{2b}$ counts the number of pairs of sets $Tsubset B$, $Xsubset Ncup B$ s.t. $|X|=2b$ and $Xcap T=varnothing$. (Indeed, one summand corresponds to choosing first $(b-j)$-element set $Tsubset B$ and then $2b$-element set $Xsubset Ncup(Bsetminus T)$.) And LHS counts such pairs with signs $(-1)^{|B setminus T|}$.



            Now there is a sign-reversing involution: find the smallest element of $Bsetminus X$ and then add it to $T$ if it's not already in $T$, otherwise remove it from $T$. So in LHS everything with non-empty $Bsetminus X$ cancels out; and there are exactly $binom nb$ variants with $Xsupset B$ and they are counted with sign $(-1)^b$.



            (And the RHS is, in fact, $(-1)^bbinom nb$.)






            share|cite|improve this answer



















            • 1




              I guess this argument also applies to the generalization $sumlimits_{j=0}^b left(-1right)^j dbinom{b}{j} dbinom{n+j}{d} = left(-1right)^b dbinom{n}{d-b}$ ?
              – darij grinberg
              Sep 28 '15 at 1:48










            • (Re: generalization) indeed
              – Grigory M
              Oct 1 '15 at 14:14













            up vote
            2
            down vote










            up vote
            2
            down vote









            Combinatorial proof of the second identity



            Let $B$ be a $b$-element set and let $N$ be an $n$-element set disjoint to $B$.



            The sum $sum_jbinom b{b-j}binom{n+j}{2b}$ counts the number of pairs of sets $Tsubset B$, $Xsubset Ncup B$ s.t. $|X|=2b$ and $Xcap T=varnothing$. (Indeed, one summand corresponds to choosing first $(b-j)$-element set $Tsubset B$ and then $2b$-element set $Xsubset Ncup(Bsetminus T)$.) And LHS counts such pairs with signs $(-1)^{|B setminus T|}$.



            Now there is a sign-reversing involution: find the smallest element of $Bsetminus X$ and then add it to $T$ if it's not already in $T$, otherwise remove it from $T$. So in LHS everything with non-empty $Bsetminus X$ cancels out; and there are exactly $binom nb$ variants with $Xsupset B$ and they are counted with sign $(-1)^b$.



            (And the RHS is, in fact, $(-1)^bbinom nb$.)






            share|cite|improve this answer














            Combinatorial proof of the second identity



            Let $B$ be a $b$-element set and let $N$ be an $n$-element set disjoint to $B$.



            The sum $sum_jbinom b{b-j}binom{n+j}{2b}$ counts the number of pairs of sets $Tsubset B$, $Xsubset Ncup B$ s.t. $|X|=2b$ and $Xcap T=varnothing$. (Indeed, one summand corresponds to choosing first $(b-j)$-element set $Tsubset B$ and then $2b$-element set $Xsubset Ncup(Bsetminus T)$.) And LHS counts such pairs with signs $(-1)^{|B setminus T|}$.



            Now there is a sign-reversing involution: find the smallest element of $Bsetminus X$ and then add it to $T$ if it's not already in $T$, otherwise remove it from $T$. So in LHS everything with non-empty $Bsetminus X$ cancels out; and there are exactly $binom nb$ variants with $Xsupset B$ and they are counted with sign $(-1)^b$.



            (And the RHS is, in fact, $(-1)^bbinom nb$.)







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 23 at 18:33









            darij grinberg

            10.1k32961




            10.1k32961










            answered Jun 20 '15 at 20:02









            Grigory M

            13.5k356103




            13.5k356103








            • 1




              I guess this argument also applies to the generalization $sumlimits_{j=0}^b left(-1right)^j dbinom{b}{j} dbinom{n+j}{d} = left(-1right)^b dbinom{n}{d-b}$ ?
              – darij grinberg
              Sep 28 '15 at 1:48










            • (Re: generalization) indeed
              – Grigory M
              Oct 1 '15 at 14:14














            • 1




              I guess this argument also applies to the generalization $sumlimits_{j=0}^b left(-1right)^j dbinom{b}{j} dbinom{n+j}{d} = left(-1right)^b dbinom{n}{d-b}$ ?
              – darij grinberg
              Sep 28 '15 at 1:48










            • (Re: generalization) indeed
              – Grigory M
              Oct 1 '15 at 14:14








            1




            1




            I guess this argument also applies to the generalization $sumlimits_{j=0}^b left(-1right)^j dbinom{b}{j} dbinom{n+j}{d} = left(-1right)^b dbinom{n}{d-b}$ ?
            – darij grinberg
            Sep 28 '15 at 1:48




            I guess this argument also applies to the generalization $sumlimits_{j=0}^b left(-1right)^j dbinom{b}{j} dbinom{n+j}{d} = left(-1right)^b dbinom{n}{d-b}$ ?
            – darij grinberg
            Sep 28 '15 at 1:48












            (Re: generalization) indeed
            – Grigory M
            Oct 1 '15 at 14:14




            (Re: generalization) indeed
            – Grigory M
            Oct 1 '15 at 14:14


















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