Produce formula












1














Let $a_{i,j}$ any positive integers wher $1leq i,j leq 3$.



I want to write closed form for following summation:



$$a_{1,3}(a_{2,1}+a_{2, 2}+a_{2,3})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{2,3}(a_{1,1}+a_{1, 2})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{3,3}(a_{1,1}+a_{1, 2})(a_{2,1}+a_{2, 2})$$



$textbf{My attempt:}$
begin{equation*}
sum_{j=1}^3 a_{j,3}left[prod_{substack{i=1 \ ineq j}}^3left(sum_{substack{k=1 \ k<j}}^{3 text{or} 3-1} a_{i,k} right) right]
end{equation*}
We will take $3-1$ when $k<j$.




Actually, I want to generalize this. My attempt so complicated to me.



If we consider all of $a_{i,j}$ as a matrix:
begin{pmatrix}
a_{1,1} & a_{1,2} & a_{1,3} \
a_{2,1} & a_{2,2} & a_{2,3} \
a_{3,1} & a_{3,2} & a_{3,3} \
end{pmatrix}
then are there any computer program to make formula?




$$a_{1,3}(a_{2,1}+a_{2, 2}+ a_{2,3})(a_{3,1}+a_{3, 2}+a_{3,3})cdots (a_{r,1}+a_{r,2}+a_{r,3})+a_{2,3}(a_{1,1}+a_{1, 2})(a_{3,1}+a_{3, 2}+a_{3,3})+cdots (a_{r,1}+a_{r,2}+a_{r,3})+a_{3,3}(a_{1,1}+a_{1, 2})(a_{2,1}+a_{2, 2})(a_{4,1}+a_{4,2}+a_{4,3}+cdots (a_{r,1}+a_{r,2}+a_{r,3}) \ vdots \
+a_{r,3}(a_{1,1}+a_{1,2})(a_{2,1}+a_{2,2})cdots (a_{r-1,1}+a_{r-1,2})
$$






combinatorics combinations







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    1














    Let $a_{i,j}$ any positive integers wher $1leq i,j leq 3$.



    I want to write closed form for following summation:



    $$a_{1,3}(a_{2,1}+a_{2, 2}+a_{2,3})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{2,3}(a_{1,1}+a_{1, 2})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{3,3}(a_{1,1}+a_{1, 2})(a_{2,1}+a_{2, 2})$$



    $textbf{My attempt:}$
    begin{equation*}
    sum_{j=1}^3 a_{j,3}left[prod_{substack{i=1 \ ineq j}}^3left(sum_{substack{k=1 \ k<j}}^{3 text{or} 3-1} a_{i,k} right) right]
    end{equation*}
    We will take $3-1$ when $k<j$.




    Actually, I want to generalize this. My attempt so complicated to me.



    If we consider all of $a_{i,j}$ as a matrix:
    begin{pmatrix}
    a_{1,1} & a_{1,2} & a_{1,3} \
    a_{2,1} & a_{2,2} & a_{2,3} \
    a_{3,1} & a_{3,2} & a_{3,3} \
    end{pmatrix}
    then are there any computer program to make formula?




    $$a_{1,3}(a_{2,1}+a_{2, 2}+ a_{2,3})(a_{3,1}+a_{3, 2}+a_{3,3})cdots (a_{r,1}+a_{r,2}+a_{r,3})+a_{2,3}(a_{1,1}+a_{1, 2})(a_{3,1}+a_{3, 2}+a_{3,3})+cdots (a_{r,1}+a_{r,2}+a_{r,3})+a_{3,3}(a_{1,1}+a_{1, 2})(a_{2,1}+a_{2, 2})(a_{4,1}+a_{4,2}+a_{4,3}+cdots (a_{r,1}+a_{r,2}+a_{r,3}) \ vdots \
    +a_{r,3}(a_{1,1}+a_{1,2})(a_{2,1}+a_{2,2})cdots (a_{r-1,1}+a_{r-1,2})
    $$






    combinatorics combinations







    share|cite|improve this question



























      1












      1








      1







      Let $a_{i,j}$ any positive integers wher $1leq i,j leq 3$.



      I want to write closed form for following summation:



      $$a_{1,3}(a_{2,1}+a_{2, 2}+a_{2,3})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{2,3}(a_{1,1}+a_{1, 2})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{3,3}(a_{1,1}+a_{1, 2})(a_{2,1}+a_{2, 2})$$



      $textbf{My attempt:}$
      begin{equation*}
      sum_{j=1}^3 a_{j,3}left[prod_{substack{i=1 \ ineq j}}^3left(sum_{substack{k=1 \ k<j}}^{3 text{or} 3-1} a_{i,k} right) right]
      end{equation*}
      We will take $3-1$ when $k<j$.




      Actually, I want to generalize this. My attempt so complicated to me.



      If we consider all of $a_{i,j}$ as a matrix:
      begin{pmatrix}
      a_{1,1} & a_{1,2} & a_{1,3} \
      a_{2,1} & a_{2,2} & a_{2,3} \
      a_{3,1} & a_{3,2} & a_{3,3} \
      end{pmatrix}
      then are there any computer program to make formula?




      $$a_{1,3}(a_{2,1}+a_{2, 2}+ a_{2,3})(a_{3,1}+a_{3, 2}+a_{3,3})cdots (a_{r,1}+a_{r,2}+a_{r,3})+a_{2,3}(a_{1,1}+a_{1, 2})(a_{3,1}+a_{3, 2}+a_{3,3})+cdots (a_{r,1}+a_{r,2}+a_{r,3})+a_{3,3}(a_{1,1}+a_{1, 2})(a_{2,1}+a_{2, 2})(a_{4,1}+a_{4,2}+a_{4,3}+cdots (a_{r,1}+a_{r,2}+a_{r,3}) \ vdots \
      +a_{r,3}(a_{1,1}+a_{1,2})(a_{2,1}+a_{2,2})cdots (a_{r-1,1}+a_{r-1,2})
      $$






      combinatorics combinations







      share|cite|improve this question















      Let $a_{i,j}$ any positive integers wher $1leq i,j leq 3$.



      I want to write closed form for following summation:



      $$a_{1,3}(a_{2,1}+a_{2, 2}+a_{2,3})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{2,3}(a_{1,1}+a_{1, 2})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{3,3}(a_{1,1}+a_{1, 2})(a_{2,1}+a_{2, 2})$$



      $textbf{My attempt:}$
      begin{equation*}
      sum_{j=1}^3 a_{j,3}left[prod_{substack{i=1 \ ineq j}}^3left(sum_{substack{k=1 \ k<j}}^{3 text{or} 3-1} a_{i,k} right) right]
      end{equation*}
      We will take $3-1$ when $k<j$.




      Actually, I want to generalize this. My attempt so complicated to me.



      If we consider all of $a_{i,j}$ as a matrix:
      begin{pmatrix}
      a_{1,1} & a_{1,2} & a_{1,3} \
      a_{2,1} & a_{2,2} & a_{2,3} \
      a_{3,1} & a_{3,2} & a_{3,3} \
      end{pmatrix}
      then are there any computer program to make formula?




      $$a_{1,3}(a_{2,1}+a_{2, 2}+ a_{2,3})(a_{3,1}+a_{3, 2}+a_{3,3})cdots (a_{r,1}+a_{r,2}+a_{r,3})+a_{2,3}(a_{1,1}+a_{1, 2})(a_{3,1}+a_{3, 2}+a_{3,3})+cdots (a_{r,1}+a_{r,2}+a_{r,3})+a_{3,3}(a_{1,1}+a_{1, 2})(a_{2,1}+a_{2, 2})(a_{4,1}+a_{4,2}+a_{4,3}+cdots (a_{r,1}+a_{r,2}+a_{r,3}) \ vdots \
      +a_{r,3}(a_{1,1}+a_{1,2})(a_{2,1}+a_{2,2})cdots (a_{r-1,1}+a_{r-1,2})
      $$






      combinatorics combinations




      combinatorics combinations






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      edited Nov 27 at 13:23

























      asked Nov 27 at 10:57









      1Spectre1

      999




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          $$a_{1,3}(a_{2,1}+a_{2, 2}+a_{2,3})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{2,3}(a_{1,1}+a_{1, 2})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{3,3}(a_{1,1}+a_{1, 2})(a_{2,1}+a_{2, 2})$$
          $$=(a_{1,1}+a_{1,2}+a_{1,3})(a_{2,1}+a_{2,2}+a_{2,3})(a_{3,1}+a_{3,2}+a_{3,3})-(a_{1,1}+a_{1,2})(a_{2,1}+a_{2,2})(a_{3,1}+a_{3,2})$$
          $$=prod_{i=1}^3sum_{j=1}^3a_{i,j}-prod_{i=1}^3sum_{j=1}^2a_{i,j}.$$
          More generally,
          $$sum_{i=1}^ra_{i,m}left(prod_{h=1}^{i-1}sum_{j=1}^{m-1}a_{h,j}right)left(prod_{h=i+1}^rsum_{j=1}^ma_{h,j}right)=prod_{i=1}^rsum_{j=1}^ma_{i,j}-prod_{i=1}^rsum_{j=1}^{m-1}a_{i,j}$$
          $$=sum_{min{j_1,j_2,dots,j_r}}a_{1,j_1}a_{2,j_2}cdots a_{r,j_r}.$$






          share|cite|improve this answer























          • Thank you for the answer. Is it possible to generalize it? I edited my question.
            – 1Spectre1
            Nov 27 at 13:24






          • 1




            @1ENİGMA1 I cleaned up my answer.
            – bof
            Nov 28 at 23:13






          • 1




            For example, $n!=prod_{i=1}^ni$ and in particular $0!=prod_{i=1}^0i=1$. Likewise, $x^n=prod_{i=1}^nx$, in particular, $x^0=prod_{i=1}^0x=1$.
            – bof
            Nov 29 at 8:59








          • 1




            $sum_{min{j_1,j_2,dots,j_r}}a_{1,j_1}a_{2,j_2}cdots a_{r,j_r}$ is the sum of all products of the form $a_{1,j_1}a_{2,j_2}cdots a_{r,j_r}$ where at least one of the indices $j_1,j_2,dots,j_r$ is equal to $m$, or in other words, $m$ is an element of the set ${j_1,j_2,dots,j_r}$.
            – bof
            Nov 29 at 9:04






          • 1




            $$prod_{i=1}^2sum_{j=1}^2a_{i,j}=(a_{1,1}+a_{1,2})(a_{2,1}+a_{2,2})=a_{1,1}a_{2,1}+a_{1,1}a_{2,2}+a_{1,2}a_{2,1}+a_{1,2}a_{2,2}$$ while $$sum_{2in{j_1,j_2}}a_{1,j_1}a_{2,j_2}=a_{1,1}a_{2,2}+a_{1,2}a_{2,1}+a_{1,2}a_{2,2}.$$ The term $a_{1,1}a_{2,1}$ is gone because $2notin{j_1,j_2}={1,1}$.
            – bof
            Nov 30 at 6:57











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          $$a_{1,3}(a_{2,1}+a_{2, 2}+a_{2,3})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{2,3}(a_{1,1}+a_{1, 2})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{3,3}(a_{1,1}+a_{1, 2})(a_{2,1}+a_{2, 2})$$
          $$=(a_{1,1}+a_{1,2}+a_{1,3})(a_{2,1}+a_{2,2}+a_{2,3})(a_{3,1}+a_{3,2}+a_{3,3})-(a_{1,1}+a_{1,2})(a_{2,1}+a_{2,2})(a_{3,1}+a_{3,2})$$
          $$=prod_{i=1}^3sum_{j=1}^3a_{i,j}-prod_{i=1}^3sum_{j=1}^2a_{i,j}.$$
          More generally,
          $$sum_{i=1}^ra_{i,m}left(prod_{h=1}^{i-1}sum_{j=1}^{m-1}a_{h,j}right)left(prod_{h=i+1}^rsum_{j=1}^ma_{h,j}right)=prod_{i=1}^rsum_{j=1}^ma_{i,j}-prod_{i=1}^rsum_{j=1}^{m-1}a_{i,j}$$
          $$=sum_{min{j_1,j_2,dots,j_r}}a_{1,j_1}a_{2,j_2}cdots a_{r,j_r}.$$






          share|cite|improve this answer























          • Thank you for the answer. Is it possible to generalize it? I edited my question.
            – 1Spectre1
            Nov 27 at 13:24






          • 1




            @1ENİGMA1 I cleaned up my answer.
            – bof
            Nov 28 at 23:13






          • 1




            For example, $n!=prod_{i=1}^ni$ and in particular $0!=prod_{i=1}^0i=1$. Likewise, $x^n=prod_{i=1}^nx$, in particular, $x^0=prod_{i=1}^0x=1$.
            – bof
            Nov 29 at 8:59








          • 1




            $sum_{min{j_1,j_2,dots,j_r}}a_{1,j_1}a_{2,j_2}cdots a_{r,j_r}$ is the sum of all products of the form $a_{1,j_1}a_{2,j_2}cdots a_{r,j_r}$ where at least one of the indices $j_1,j_2,dots,j_r$ is equal to $m$, or in other words, $m$ is an element of the set ${j_1,j_2,dots,j_r}$.
            – bof
            Nov 29 at 9:04






          • 1




            $$prod_{i=1}^2sum_{j=1}^2a_{i,j}=(a_{1,1}+a_{1,2})(a_{2,1}+a_{2,2})=a_{1,1}a_{2,1}+a_{1,1}a_{2,2}+a_{1,2}a_{2,1}+a_{1,2}a_{2,2}$$ while $$sum_{2in{j_1,j_2}}a_{1,j_1}a_{2,j_2}=a_{1,1}a_{2,2}+a_{1,2}a_{2,1}+a_{1,2}a_{2,2}.$$ The term $a_{1,1}a_{2,1}$ is gone because $2notin{j_1,j_2}={1,1}$.
            – bof
            Nov 30 at 6:57
















          3














          $$a_{1,3}(a_{2,1}+a_{2, 2}+a_{2,3})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{2,3}(a_{1,1}+a_{1, 2})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{3,3}(a_{1,1}+a_{1, 2})(a_{2,1}+a_{2, 2})$$
          $$=(a_{1,1}+a_{1,2}+a_{1,3})(a_{2,1}+a_{2,2}+a_{2,3})(a_{3,1}+a_{3,2}+a_{3,3})-(a_{1,1}+a_{1,2})(a_{2,1}+a_{2,2})(a_{3,1}+a_{3,2})$$
          $$=prod_{i=1}^3sum_{j=1}^3a_{i,j}-prod_{i=1}^3sum_{j=1}^2a_{i,j}.$$
          More generally,
          $$sum_{i=1}^ra_{i,m}left(prod_{h=1}^{i-1}sum_{j=1}^{m-1}a_{h,j}right)left(prod_{h=i+1}^rsum_{j=1}^ma_{h,j}right)=prod_{i=1}^rsum_{j=1}^ma_{i,j}-prod_{i=1}^rsum_{j=1}^{m-1}a_{i,j}$$
          $$=sum_{min{j_1,j_2,dots,j_r}}a_{1,j_1}a_{2,j_2}cdots a_{r,j_r}.$$






          share|cite|improve this answer























          • Thank you for the answer. Is it possible to generalize it? I edited my question.
            – 1Spectre1
            Nov 27 at 13:24






          • 1




            @1ENİGMA1 I cleaned up my answer.
            – bof
            Nov 28 at 23:13






          • 1




            For example, $n!=prod_{i=1}^ni$ and in particular $0!=prod_{i=1}^0i=1$. Likewise, $x^n=prod_{i=1}^nx$, in particular, $x^0=prod_{i=1}^0x=1$.
            – bof
            Nov 29 at 8:59








          • 1




            $sum_{min{j_1,j_2,dots,j_r}}a_{1,j_1}a_{2,j_2}cdots a_{r,j_r}$ is the sum of all products of the form $a_{1,j_1}a_{2,j_2}cdots a_{r,j_r}$ where at least one of the indices $j_1,j_2,dots,j_r$ is equal to $m$, or in other words, $m$ is an element of the set ${j_1,j_2,dots,j_r}$.
            – bof
            Nov 29 at 9:04






          • 1




            $$prod_{i=1}^2sum_{j=1}^2a_{i,j}=(a_{1,1}+a_{1,2})(a_{2,1}+a_{2,2})=a_{1,1}a_{2,1}+a_{1,1}a_{2,2}+a_{1,2}a_{2,1}+a_{1,2}a_{2,2}$$ while $$sum_{2in{j_1,j_2}}a_{1,j_1}a_{2,j_2}=a_{1,1}a_{2,2}+a_{1,2}a_{2,1}+a_{1,2}a_{2,2}.$$ The term $a_{1,1}a_{2,1}$ is gone because $2notin{j_1,j_2}={1,1}$.
            – bof
            Nov 30 at 6:57














          3












          3








          3






          $$a_{1,3}(a_{2,1}+a_{2, 2}+a_{2,3})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{2,3}(a_{1,1}+a_{1, 2})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{3,3}(a_{1,1}+a_{1, 2})(a_{2,1}+a_{2, 2})$$
          $$=(a_{1,1}+a_{1,2}+a_{1,3})(a_{2,1}+a_{2,2}+a_{2,3})(a_{3,1}+a_{3,2}+a_{3,3})-(a_{1,1}+a_{1,2})(a_{2,1}+a_{2,2})(a_{3,1}+a_{3,2})$$
          $$=prod_{i=1}^3sum_{j=1}^3a_{i,j}-prod_{i=1}^3sum_{j=1}^2a_{i,j}.$$
          More generally,
          $$sum_{i=1}^ra_{i,m}left(prod_{h=1}^{i-1}sum_{j=1}^{m-1}a_{h,j}right)left(prod_{h=i+1}^rsum_{j=1}^ma_{h,j}right)=prod_{i=1}^rsum_{j=1}^ma_{i,j}-prod_{i=1}^rsum_{j=1}^{m-1}a_{i,j}$$
          $$=sum_{min{j_1,j_2,dots,j_r}}a_{1,j_1}a_{2,j_2}cdots a_{r,j_r}.$$






          share|cite|improve this answer














          $$a_{1,3}(a_{2,1}+a_{2, 2}+a_{2,3})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{2,3}(a_{1,1}+a_{1, 2})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{3,3}(a_{1,1}+a_{1, 2})(a_{2,1}+a_{2, 2})$$
          $$=(a_{1,1}+a_{1,2}+a_{1,3})(a_{2,1}+a_{2,2}+a_{2,3})(a_{3,1}+a_{3,2}+a_{3,3})-(a_{1,1}+a_{1,2})(a_{2,1}+a_{2,2})(a_{3,1}+a_{3,2})$$
          $$=prod_{i=1}^3sum_{j=1}^3a_{i,j}-prod_{i=1}^3sum_{j=1}^2a_{i,j}.$$
          More generally,
          $$sum_{i=1}^ra_{i,m}left(prod_{h=1}^{i-1}sum_{j=1}^{m-1}a_{h,j}right)left(prod_{h=i+1}^rsum_{j=1}^ma_{h,j}right)=prod_{i=1}^rsum_{j=1}^ma_{i,j}-prod_{i=1}^rsum_{j=1}^{m-1}a_{i,j}$$
          $$=sum_{min{j_1,j_2,dots,j_r}}a_{1,j_1}a_{2,j_2}cdots a_{r,j_r}.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 28 at 23:11

























          answered Nov 27 at 12:27









          bof

          50.1k457119




          50.1k457119












          • Thank you for the answer. Is it possible to generalize it? I edited my question.
            – 1Spectre1
            Nov 27 at 13:24






          • 1




            @1ENİGMA1 I cleaned up my answer.
            – bof
            Nov 28 at 23:13






          • 1




            For example, $n!=prod_{i=1}^ni$ and in particular $0!=prod_{i=1}^0i=1$. Likewise, $x^n=prod_{i=1}^nx$, in particular, $x^0=prod_{i=1}^0x=1$.
            – bof
            Nov 29 at 8:59








          • 1




            $sum_{min{j_1,j_2,dots,j_r}}a_{1,j_1}a_{2,j_2}cdots a_{r,j_r}$ is the sum of all products of the form $a_{1,j_1}a_{2,j_2}cdots a_{r,j_r}$ where at least one of the indices $j_1,j_2,dots,j_r$ is equal to $m$, or in other words, $m$ is an element of the set ${j_1,j_2,dots,j_r}$.
            – bof
            Nov 29 at 9:04






          • 1




            $$prod_{i=1}^2sum_{j=1}^2a_{i,j}=(a_{1,1}+a_{1,2})(a_{2,1}+a_{2,2})=a_{1,1}a_{2,1}+a_{1,1}a_{2,2}+a_{1,2}a_{2,1}+a_{1,2}a_{2,2}$$ while $$sum_{2in{j_1,j_2}}a_{1,j_1}a_{2,j_2}=a_{1,1}a_{2,2}+a_{1,2}a_{2,1}+a_{1,2}a_{2,2}.$$ The term $a_{1,1}a_{2,1}$ is gone because $2notin{j_1,j_2}={1,1}$.
            – bof
            Nov 30 at 6:57


















          • Thank you for the answer. Is it possible to generalize it? I edited my question.
            – 1Spectre1
            Nov 27 at 13:24






          • 1




            @1ENİGMA1 I cleaned up my answer.
            – bof
            Nov 28 at 23:13






          • 1




            For example, $n!=prod_{i=1}^ni$ and in particular $0!=prod_{i=1}^0i=1$. Likewise, $x^n=prod_{i=1}^nx$, in particular, $x^0=prod_{i=1}^0x=1$.
            – bof
            Nov 29 at 8:59








          • 1




            $sum_{min{j_1,j_2,dots,j_r}}a_{1,j_1}a_{2,j_2}cdots a_{r,j_r}$ is the sum of all products of the form $a_{1,j_1}a_{2,j_2}cdots a_{r,j_r}$ where at least one of the indices $j_1,j_2,dots,j_r$ is equal to $m$, or in other words, $m$ is an element of the set ${j_1,j_2,dots,j_r}$.
            – bof
            Nov 29 at 9:04






          • 1




            $$prod_{i=1}^2sum_{j=1}^2a_{i,j}=(a_{1,1}+a_{1,2})(a_{2,1}+a_{2,2})=a_{1,1}a_{2,1}+a_{1,1}a_{2,2}+a_{1,2}a_{2,1}+a_{1,2}a_{2,2}$$ while $$sum_{2in{j_1,j_2}}a_{1,j_1}a_{2,j_2}=a_{1,1}a_{2,2}+a_{1,2}a_{2,1}+a_{1,2}a_{2,2}.$$ The term $a_{1,1}a_{2,1}$ is gone because $2notin{j_1,j_2}={1,1}$.
            – bof
            Nov 30 at 6:57
















          Thank you for the answer. Is it possible to generalize it? I edited my question.
          – 1Spectre1
          Nov 27 at 13:24




          Thank you for the answer. Is it possible to generalize it? I edited my question.
          – 1Spectre1
          Nov 27 at 13:24




          1




          1




          @1ENİGMA1 I cleaned up my answer.
          – bof
          Nov 28 at 23:13




          @1ENİGMA1 I cleaned up my answer.
          – bof
          Nov 28 at 23:13




          1




          1




          For example, $n!=prod_{i=1}^ni$ and in particular $0!=prod_{i=1}^0i=1$. Likewise, $x^n=prod_{i=1}^nx$, in particular, $x^0=prod_{i=1}^0x=1$.
          – bof
          Nov 29 at 8:59






          For example, $n!=prod_{i=1}^ni$ and in particular $0!=prod_{i=1}^0i=1$. Likewise, $x^n=prod_{i=1}^nx$, in particular, $x^0=prod_{i=1}^0x=1$.
          – bof
          Nov 29 at 8:59






          1




          1




          $sum_{min{j_1,j_2,dots,j_r}}a_{1,j_1}a_{2,j_2}cdots a_{r,j_r}$ is the sum of all products of the form $a_{1,j_1}a_{2,j_2}cdots a_{r,j_r}$ where at least one of the indices $j_1,j_2,dots,j_r$ is equal to $m$, or in other words, $m$ is an element of the set ${j_1,j_2,dots,j_r}$.
          – bof
          Nov 29 at 9:04




          $sum_{min{j_1,j_2,dots,j_r}}a_{1,j_1}a_{2,j_2}cdots a_{r,j_r}$ is the sum of all products of the form $a_{1,j_1}a_{2,j_2}cdots a_{r,j_r}$ where at least one of the indices $j_1,j_2,dots,j_r$ is equal to $m$, or in other words, $m$ is an element of the set ${j_1,j_2,dots,j_r}$.
          – bof
          Nov 29 at 9:04




          1




          1




          $$prod_{i=1}^2sum_{j=1}^2a_{i,j}=(a_{1,1}+a_{1,2})(a_{2,1}+a_{2,2})=a_{1,1}a_{2,1}+a_{1,1}a_{2,2}+a_{1,2}a_{2,1}+a_{1,2}a_{2,2}$$ while $$sum_{2in{j_1,j_2}}a_{1,j_1}a_{2,j_2}=a_{1,1}a_{2,2}+a_{1,2}a_{2,1}+a_{1,2}a_{2,2}.$$ The term $a_{1,1}a_{2,1}$ is gone because $2notin{j_1,j_2}={1,1}$.
          – bof
          Nov 30 at 6:57




          $$prod_{i=1}^2sum_{j=1}^2a_{i,j}=(a_{1,1}+a_{1,2})(a_{2,1}+a_{2,2})=a_{1,1}a_{2,1}+a_{1,1}a_{2,2}+a_{1,2}a_{2,1}+a_{1,2}a_{2,2}$$ while $$sum_{2in{j_1,j_2}}a_{1,j_1}a_{2,j_2}=a_{1,1}a_{2,2}+a_{1,2}a_{2,1}+a_{1,2}a_{2,2}.$$ The term $a_{1,1}a_{2,1}$ is gone because $2notin{j_1,j_2}={1,1}$.
          – bof
          Nov 30 at 6:57


















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