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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