Find a Closed form for the Combinatorial Sum $sum_{k=0}^mbinom{n-k}{m-k} $ and Provide a Combinatorial Proof...












2












$begingroup$



Question
Find a closed form for the combinatorial sum $sum_{k=0}^mbinom{n-k}{m-k} $ and provide a combinatorial proof of the resulting identity.




My attempt



I was able to find a closed form using the method of "snake-oil" but unable to provide a combinatorial proof. We claim that
$$
sum_{k=0}^mbinom{n-k}{m-k}=binom{n+1}{m}tag{0}.
$$
Indeed note that (formally)
$$
begin{align}
sum_{m=0}^inftyleft(sum_{k=0}^mbinom{n-k}{m-k}right)z^m&=sum_{k=0}^inftyleft(sum_{m=k}^inftybinom{n-k}{m-k}right)z^mtag{1}\
&=sum_{k=0}^infty z^kleft(sum_{u=0}^inftybinom{n-k}{u}z^uright)\
&=sum_{k=0}^infty z^k(1+z)^{n-k}tag{2}\
&=(1+z)^nsum_{k=0}^inftyleft(frac{z}{1+z}right)^k\
&=(1+z)^nfrac{1}{1-frac{z}{1+z}}=(1+z)^{n+1}.tag{3}
end{align}
$$
Taking the coefficient of $z^m$ from $(1+z)^{n+1}$ yields the result. In the above computation we interchanged summation in $(1)$, used the binomial thoerem in $(2)$ and used the formula for a geometric series in $(3)$.



My problem



The simplicity of the identity in $(0)$ (supposing I have not made any mistakes) suggests a combinatorial proof. Unfortunately, I have not been able to make much progress here. I don't know how to classify the $m$ element subsets of $[n+1]$ to obtain $(0)$.



Any help is appreciated.






combinatorics discrete-mathematics binomial-coefficients generating-functions







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    If you replace $m-k$ with $n-m$ on the left, and the $m$ on the right with $n-m$, then this becomes the hockey stick identity.
    $endgroup$
    – Mike Earnest
    Dec 25 '18 at 17:18










  • $begingroup$
    Partition the set of subsets of ${1,2,ldots,n+1}$ of size $m$ into $m+1$ parts $P_i$, $iin{0,1,ldots,m}$ where $P_i$ is the set of size-$m$ subsets whose least missing element is $i+1$.
    $endgroup$
    – Will Orrick
    Dec 26 '18 at 1:05


















2












$begingroup$



Question
Find a closed form for the combinatorial sum $sum_{k=0}^mbinom{n-k}{m-k} $ and provide a combinatorial proof of the resulting identity.




My attempt



I was able to find a closed form using the method of "snake-oil" but unable to provide a combinatorial proof. We claim that
$$
sum_{k=0}^mbinom{n-k}{m-k}=binom{n+1}{m}tag{0}.
$$
Indeed note that (formally)
$$
begin{align}
sum_{m=0}^inftyleft(sum_{k=0}^mbinom{n-k}{m-k}right)z^m&=sum_{k=0}^inftyleft(sum_{m=k}^inftybinom{n-k}{m-k}right)z^mtag{1}\
&=sum_{k=0}^infty z^kleft(sum_{u=0}^inftybinom{n-k}{u}z^uright)\
&=sum_{k=0}^infty z^k(1+z)^{n-k}tag{2}\
&=(1+z)^nsum_{k=0}^inftyleft(frac{z}{1+z}right)^k\
&=(1+z)^nfrac{1}{1-frac{z}{1+z}}=(1+z)^{n+1}.tag{3}
end{align}
$$
Taking the coefficient of $z^m$ from $(1+z)^{n+1}$ yields the result. In the above computation we interchanged summation in $(1)$, used the binomial thoerem in $(2)$ and used the formula for a geometric series in $(3)$.



My problem



The simplicity of the identity in $(0)$ (supposing I have not made any mistakes) suggests a combinatorial proof. Unfortunately, I have not been able to make much progress here. I don't know how to classify the $m$ element subsets of $[n+1]$ to obtain $(0)$.



Any help is appreciated.






combinatorics discrete-mathematics binomial-coefficients generating-functions







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    If you replace $m-k$ with $n-m$ on the left, and the $m$ on the right with $n-m$, then this becomes the hockey stick identity.
    $endgroup$
    – Mike Earnest
    Dec 25 '18 at 17:18










  • $begingroup$
    Partition the set of subsets of ${1,2,ldots,n+1}$ of size $m$ into $m+1$ parts $P_i$, $iin{0,1,ldots,m}$ where $P_i$ is the set of size-$m$ subsets whose least missing element is $i+1$.
    $endgroup$
    – Will Orrick
    Dec 26 '18 at 1:05
















2












2








2





$begingroup$



Question
Find a closed form for the combinatorial sum $sum_{k=0}^mbinom{n-k}{m-k} $ and provide a combinatorial proof of the resulting identity.




My attempt



I was able to find a closed form using the method of "snake-oil" but unable to provide a combinatorial proof. We claim that
$$
sum_{k=0}^mbinom{n-k}{m-k}=binom{n+1}{m}tag{0}.
$$
Indeed note that (formally)
$$
begin{align}
sum_{m=0}^inftyleft(sum_{k=0}^mbinom{n-k}{m-k}right)z^m&=sum_{k=0}^inftyleft(sum_{m=k}^inftybinom{n-k}{m-k}right)z^mtag{1}\
&=sum_{k=0}^infty z^kleft(sum_{u=0}^inftybinom{n-k}{u}z^uright)\
&=sum_{k=0}^infty z^k(1+z)^{n-k}tag{2}\
&=(1+z)^nsum_{k=0}^inftyleft(frac{z}{1+z}right)^k\
&=(1+z)^nfrac{1}{1-frac{z}{1+z}}=(1+z)^{n+1}.tag{3}
end{align}
$$
Taking the coefficient of $z^m$ from $(1+z)^{n+1}$ yields the result. In the above computation we interchanged summation in $(1)$, used the binomial thoerem in $(2)$ and used the formula for a geometric series in $(3)$.



My problem



The simplicity of the identity in $(0)$ (supposing I have not made any mistakes) suggests a combinatorial proof. Unfortunately, I have not been able to make much progress here. I don't know how to classify the $m$ element subsets of $[n+1]$ to obtain $(0)$.



Any help is appreciated.






combinatorics discrete-mathematics binomial-coefficients generating-functions







share|cite|improve this question









$endgroup$





Question
Find a closed form for the combinatorial sum $sum_{k=0}^mbinom{n-k}{m-k} $ and provide a combinatorial proof of the resulting identity.




My attempt



I was able to find a closed form using the method of "snake-oil" but unable to provide a combinatorial proof. We claim that
$$
sum_{k=0}^mbinom{n-k}{m-k}=binom{n+1}{m}tag{0}.
$$
Indeed note that (formally)
$$
begin{align}
sum_{m=0}^inftyleft(sum_{k=0}^mbinom{n-k}{m-k}right)z^m&=sum_{k=0}^inftyleft(sum_{m=k}^inftybinom{n-k}{m-k}right)z^mtag{1}\
&=sum_{k=0}^infty z^kleft(sum_{u=0}^inftybinom{n-k}{u}z^uright)\
&=sum_{k=0}^infty z^k(1+z)^{n-k}tag{2}\
&=(1+z)^nsum_{k=0}^inftyleft(frac{z}{1+z}right)^k\
&=(1+z)^nfrac{1}{1-frac{z}{1+z}}=(1+z)^{n+1}.tag{3}
end{align}
$$
Taking the coefficient of $z^m$ from $(1+z)^{n+1}$ yields the result. In the above computation we interchanged summation in $(1)$, used the binomial thoerem in $(2)$ and used the formula for a geometric series in $(3)$.



My problem



The simplicity of the identity in $(0)$ (supposing I have not made any mistakes) suggests a combinatorial proof. Unfortunately, I have not been able to make much progress here. I don't know how to classify the $m$ element subsets of $[n+1]$ to obtain $(0)$.



Any help is appreciated.






combinatorics discrete-mathematics binomial-coefficients generating-functions




combinatorics discrete-mathematics binomial-coefficients generating-functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 25 '18 at 17:07









Foobaz JohnFoobaz John

22.1k41452




22.1k41452








  • 3




    $begingroup$
    If you replace $m-k$ with $n-m$ on the left, and the $m$ on the right with $n-m$, then this becomes the hockey stick identity.
    $endgroup$
    – Mike Earnest
    Dec 25 '18 at 17:18










  • $begingroup$
    Partition the set of subsets of ${1,2,ldots,n+1}$ of size $m$ into $m+1$ parts $P_i$, $iin{0,1,ldots,m}$ where $P_i$ is the set of size-$m$ subsets whose least missing element is $i+1$.
    $endgroup$
    – Will Orrick
    Dec 26 '18 at 1:05
















  • 3




    $begingroup$
    If you replace $m-k$ with $n-m$ on the left, and the $m$ on the right with $n-m$, then this becomes the hockey stick identity.
    $endgroup$
    – Mike Earnest
    Dec 25 '18 at 17:18










  • $begingroup$
    Partition the set of subsets of ${1,2,ldots,n+1}$ of size $m$ into $m+1$ parts $P_i$, $iin{0,1,ldots,m}$ where $P_i$ is the set of size-$m$ subsets whose least missing element is $i+1$.
    $endgroup$
    – Will Orrick
    Dec 26 '18 at 1:05










3




3




$begingroup$
If you replace $m-k$ with $n-m$ on the left, and the $m$ on the right with $n-m$, then this becomes the hockey stick identity.
$endgroup$
– Mike Earnest
Dec 25 '18 at 17:18




$begingroup$
If you replace $m-k$ with $n-m$ on the left, and the $m$ on the right with $n-m$, then this becomes the hockey stick identity.
$endgroup$
– Mike Earnest
Dec 25 '18 at 17:18












$begingroup$
Partition the set of subsets of ${1,2,ldots,n+1}$ of size $m$ into $m+1$ parts $P_i$, $iin{0,1,ldots,m}$ where $P_i$ is the set of size-$m$ subsets whose least missing element is $i+1$.
$endgroup$
– Will Orrick
Dec 26 '18 at 1:05






$begingroup$
Partition the set of subsets of ${1,2,ldots,n+1}$ of size $m$ into $m+1$ parts $P_i$, $iin{0,1,ldots,m}$ where $P_i$ is the set of size-$m$ subsets whose least missing element is $i+1$.
$endgroup$
– Will Orrick
Dec 26 '18 at 1:05












2 Answers
2






active

oldest

votes


















2












$begingroup$

In how many ways can you choose the m-element subsets of {1,...,n}? In ${n choose m}$ ways. But you may also write this as (the number of ways to choose m-element subsets that contain n )+ (the ones that don’t contain n but contain (n-1) +(the number of subsets that contain neither n, nor n-1, but contain n-2) +...(keep going)+(the number of subsets that contain neither n, nor n-1,nor...,nor m+1) and this gives you ${n-1 choose m} + {n-2 choose m}+...+{m choose m}$



This is the same identity as the one you need to prove once you note that ${n-k choose m-k}={n-k choose n-m}$ and change the order of sumation from m to 0 instead of 0 to m.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Lemma:




    $$binom{n}{k}=binom{n}{n-k}$$




    $$binom{n}{k}=frac{n!}{(n-k)!k!}=binom{n}{n-k}=frac{n!}{k!(n-k)!}$$
    Proof:



    $$binom{n-k}{m-k}=binom{n-k}{(n-k)-(m-k)}=binom{n-k}{n-m}$$
    Therefore,
    $$sum_{k=0}^{m}binom{n-k}{m-k}=sum_{k=0}^{m}binom{n-k}{n-m}$$
    Notice that
    $$sum_{k=0}^{m}binom{n-k}{n-m}=sum_{k=n-m}^{n}binom{k}{n-m}$$
    Using Hockey Stick identity
    $$sum_{i=r}^{n}binom{i}{r}=binom{n+1}{r+1}$$
    We have
    $$sum_{k=0}^{m}binom{n-k}{m-k}=sum_{k=0}^{m}binom{n-k}{n-m}=sum_{k=n-m}^{n}binom{k}{n-m}=binom{n+1}{n+1-m}=binom{n+1}{m}$$






    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









      2












      $begingroup$

      In how many ways can you choose the m-element subsets of {1,...,n}? In ${n choose m}$ ways. But you may also write this as (the number of ways to choose m-element subsets that contain n )+ (the ones that don’t contain n but contain (n-1) +(the number of subsets that contain neither n, nor n-1, but contain n-2) +...(keep going)+(the number of subsets that contain neither n, nor n-1,nor...,nor m+1) and this gives you ${n-1 choose m} + {n-2 choose m}+...+{m choose m}$



      This is the same identity as the one you need to prove once you note that ${n-k choose m-k}={n-k choose n-m}$ and change the order of sumation from m to 0 instead of 0 to m.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        In how many ways can you choose the m-element subsets of {1,...,n}? In ${n choose m}$ ways. But you may also write this as (the number of ways to choose m-element subsets that contain n )+ (the ones that don’t contain n but contain (n-1) +(the number of subsets that contain neither n, nor n-1, but contain n-2) +...(keep going)+(the number of subsets that contain neither n, nor n-1,nor...,nor m+1) and this gives you ${n-1 choose m} + {n-2 choose m}+...+{m choose m}$



        This is the same identity as the one you need to prove once you note that ${n-k choose m-k}={n-k choose n-m}$ and change the order of sumation from m to 0 instead of 0 to m.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          In how many ways can you choose the m-element subsets of {1,...,n}? In ${n choose m}$ ways. But you may also write this as (the number of ways to choose m-element subsets that contain n )+ (the ones that don’t contain n but contain (n-1) +(the number of subsets that contain neither n, nor n-1, but contain n-2) +...(keep going)+(the number of subsets that contain neither n, nor n-1,nor...,nor m+1) and this gives you ${n-1 choose m} + {n-2 choose m}+...+{m choose m}$



          This is the same identity as the one you need to prove once you note that ${n-k choose m-k}={n-k choose n-m}$ and change the order of sumation from m to 0 instead of 0 to m.






          share|cite|improve this answer











          $endgroup$



          In how many ways can you choose the m-element subsets of {1,...,n}? In ${n choose m}$ ways. But you may also write this as (the number of ways to choose m-element subsets that contain n )+ (the ones that don’t contain n but contain (n-1) +(the number of subsets that contain neither n, nor n-1, but contain n-2) +...(keep going)+(the number of subsets that contain neither n, nor n-1,nor...,nor m+1) and this gives you ${n-1 choose m} + {n-2 choose m}+...+{m choose m}$



          This is the same identity as the one you need to prove once you note that ${n-k choose m-k}={n-k choose n-m}$ and change the order of sumation from m to 0 instead of 0 to m.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 25 '18 at 17:52

























          answered Dec 25 '18 at 17:21









          Sorin TircSorin Tirc

          1,800213




          1,800213























              0












              $begingroup$

              Lemma:




              $$binom{n}{k}=binom{n}{n-k}$$




              $$binom{n}{k}=frac{n!}{(n-k)!k!}=binom{n}{n-k}=frac{n!}{k!(n-k)!}$$
              Proof:



              $$binom{n-k}{m-k}=binom{n-k}{(n-k)-(m-k)}=binom{n-k}{n-m}$$
              Therefore,
              $$sum_{k=0}^{m}binom{n-k}{m-k}=sum_{k=0}^{m}binom{n-k}{n-m}$$
              Notice that
              $$sum_{k=0}^{m}binom{n-k}{n-m}=sum_{k=n-m}^{n}binom{k}{n-m}$$
              Using Hockey Stick identity
              $$sum_{i=r}^{n}binom{i}{r}=binom{n+1}{r+1}$$
              We have
              $$sum_{k=0}^{m}binom{n-k}{m-k}=sum_{k=0}^{m}binom{n-k}{n-m}=sum_{k=n-m}^{n}binom{k}{n-m}=binom{n+1}{n+1-m}=binom{n+1}{m}$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Lemma:




                $$binom{n}{k}=binom{n}{n-k}$$




                $$binom{n}{k}=frac{n!}{(n-k)!k!}=binom{n}{n-k}=frac{n!}{k!(n-k)!}$$
                Proof:



                $$binom{n-k}{m-k}=binom{n-k}{(n-k)-(m-k)}=binom{n-k}{n-m}$$
                Therefore,
                $$sum_{k=0}^{m}binom{n-k}{m-k}=sum_{k=0}^{m}binom{n-k}{n-m}$$
                Notice that
                $$sum_{k=0}^{m}binom{n-k}{n-m}=sum_{k=n-m}^{n}binom{k}{n-m}$$
                Using Hockey Stick identity
                $$sum_{i=r}^{n}binom{i}{r}=binom{n+1}{r+1}$$
                We have
                $$sum_{k=0}^{m}binom{n-k}{m-k}=sum_{k=0}^{m}binom{n-k}{n-m}=sum_{k=n-m}^{n}binom{k}{n-m}=binom{n+1}{n+1-m}=binom{n+1}{m}$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Lemma:




                  $$binom{n}{k}=binom{n}{n-k}$$




                  $$binom{n}{k}=frac{n!}{(n-k)!k!}=binom{n}{n-k}=frac{n!}{k!(n-k)!}$$
                  Proof:



                  $$binom{n-k}{m-k}=binom{n-k}{(n-k)-(m-k)}=binom{n-k}{n-m}$$
                  Therefore,
                  $$sum_{k=0}^{m}binom{n-k}{m-k}=sum_{k=0}^{m}binom{n-k}{n-m}$$
                  Notice that
                  $$sum_{k=0}^{m}binom{n-k}{n-m}=sum_{k=n-m}^{n}binom{k}{n-m}$$
                  Using Hockey Stick identity
                  $$sum_{i=r}^{n}binom{i}{r}=binom{n+1}{r+1}$$
                  We have
                  $$sum_{k=0}^{m}binom{n-k}{m-k}=sum_{k=0}^{m}binom{n-k}{n-m}=sum_{k=n-m}^{n}binom{k}{n-m}=binom{n+1}{n+1-m}=binom{n+1}{m}$$






                  share|cite|improve this answer









                  $endgroup$



                  Lemma:




                  $$binom{n}{k}=binom{n}{n-k}$$




                  $$binom{n}{k}=frac{n!}{(n-k)!k!}=binom{n}{n-k}=frac{n!}{k!(n-k)!}$$
                  Proof:



                  $$binom{n-k}{m-k}=binom{n-k}{(n-k)-(m-k)}=binom{n-k}{n-m}$$
                  Therefore,
                  $$sum_{k=0}^{m}binom{n-k}{m-k}=sum_{k=0}^{m}binom{n-k}{n-m}$$
                  Notice that
                  $$sum_{k=0}^{m}binom{n-k}{n-m}=sum_{k=n-m}^{n}binom{k}{n-m}$$
                  Using Hockey Stick identity
                  $$sum_{i=r}^{n}binom{i}{r}=binom{n+1}{r+1}$$
                  We have
                  $$sum_{k=0}^{m}binom{n-k}{m-k}=sum_{k=0}^{m}binom{n-k}{n-m}=sum_{k=n-m}^{n}binom{k}{n-m}=binom{n+1}{n+1-m}=binom{n+1}{m}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 25 '18 at 18:28









                  LarryLarry

                  2,41331129




                  2,41331129






























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                      Mont Emei

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                      Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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