How many non-negative solutions for $x_{1} + x_{2} + x_{3} + x_{4} = 40$ where $2 leq x_{1} leq 8, x_{2} leq...












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My solution:



We have:
$x_{1} + x_{2} + x_{3} + x_{4} = 40$ where $2 leq x_{1} leq 8, x_{2} leq 4, x_{3} geq 4, x_{4} leq 5$



$Leftrightarrow x_{2} + x_{3} + x_{4} = 40 - x_{1} quad (*)$



Consider:



$x_{2} + x_{3} + x_{4} = 40 - x_{1}$ where $x_{2} geq 0, x_{3} geq 4, x_{4} geq 0 quad (**)$



$x_{2} + x_{3} + x_{4} = 40 - x_{1}$ where $x_{2} geq 5, x_{3} geq 4, x_{4} geq 6 quad (***)$



Let $f$ is the function that compute the number of non-negative solutions of an equation.



$implies f(*) = f(**) - f(***)$



Thus, the number of non-negative solutions of (*) is $sum_{x_{1} = 2}^{8}( {40 - x_{1} + 3 - 1 choose 3 - 1} - {25-x_{1}+3 - 1 choose 3 - 1}) = 3045$



I found that the right answer is 210 by trying some programming script. But I don't know what was wrong with my solution. Please help me. Thank you!






combinatorics







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  • 1




    $begingroup$
    The three upper limits on $x_1, x_2, x_4$ and that the only variable not upper restrained is dependent on the three restrained makes the stars and bars/choose solution too restrained to be relevent. Do this as pure choose/multiplying. There are so many chooses an the three restrained variables, and the unrestrained can be set up to be completely dependent on the other three. So it is straight mulitplication
    $endgroup$
    – fleablood
    Dec 9 '18 at 16:39






  • 1




    $begingroup$
    YOu are way over counting. You have the solution for $x_2< 5$ OR $x_4< 5$. You don't want that you want AND. ANd you also never take $x_3 ge 4$ into account. The first term should be $36$ not $40$.
    $endgroup$
    – fleablood
    Dec 9 '18 at 16:47












  • $begingroup$
    @fleablood thank for your comment, this is my wrong.
    $endgroup$
    – Duy Huynh
    Dec 9 '18 at 16:51
















3












$begingroup$


My solution:



We have:
$x_{1} + x_{2} + x_{3} + x_{4} = 40$ where $2 leq x_{1} leq 8, x_{2} leq 4, x_{3} geq 4, x_{4} leq 5$



$Leftrightarrow x_{2} + x_{3} + x_{4} = 40 - x_{1} quad (*)$



Consider:



$x_{2} + x_{3} + x_{4} = 40 - x_{1}$ where $x_{2} geq 0, x_{3} geq 4, x_{4} geq 0 quad (**)$



$x_{2} + x_{3} + x_{4} = 40 - x_{1}$ where $x_{2} geq 5, x_{3} geq 4, x_{4} geq 6 quad (***)$



Let $f$ is the function that compute the number of non-negative solutions of an equation.



$implies f(*) = f(**) - f(***)$



Thus, the number of non-negative solutions of (*) is $sum_{x_{1} = 2}^{8}( {40 - x_{1} + 3 - 1 choose 3 - 1} - {25-x_{1}+3 - 1 choose 3 - 1}) = 3045$



I found that the right answer is 210 by trying some programming script. But I don't know what was wrong with my solution. Please help me. Thank you!






combinatorics







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The three upper limits on $x_1, x_2, x_4$ and that the only variable not upper restrained is dependent on the three restrained makes the stars and bars/choose solution too restrained to be relevent. Do this as pure choose/multiplying. There are so many chooses an the three restrained variables, and the unrestrained can be set up to be completely dependent on the other three. So it is straight mulitplication
    $endgroup$
    – fleablood
    Dec 9 '18 at 16:39






  • 1




    $begingroup$
    YOu are way over counting. You have the solution for $x_2< 5$ OR $x_4< 5$. You don't want that you want AND. ANd you also never take $x_3 ge 4$ into account. The first term should be $36$ not $40$.
    $endgroup$
    – fleablood
    Dec 9 '18 at 16:47












  • $begingroup$
    @fleablood thank for your comment, this is my wrong.
    $endgroup$
    – Duy Huynh
    Dec 9 '18 at 16:51














3












3








3


1



$begingroup$


My solution:



We have:
$x_{1} + x_{2} + x_{3} + x_{4} = 40$ where $2 leq x_{1} leq 8, x_{2} leq 4, x_{3} geq 4, x_{4} leq 5$



$Leftrightarrow x_{2} + x_{3} + x_{4} = 40 - x_{1} quad (*)$



Consider:



$x_{2} + x_{3} + x_{4} = 40 - x_{1}$ where $x_{2} geq 0, x_{3} geq 4, x_{4} geq 0 quad (**)$



$x_{2} + x_{3} + x_{4} = 40 - x_{1}$ where $x_{2} geq 5, x_{3} geq 4, x_{4} geq 6 quad (***)$



Let $f$ is the function that compute the number of non-negative solutions of an equation.



$implies f(*) = f(**) - f(***)$



Thus, the number of non-negative solutions of (*) is $sum_{x_{1} = 2}^{8}( {40 - x_{1} + 3 - 1 choose 3 - 1} - {25-x_{1}+3 - 1 choose 3 - 1}) = 3045$



I found that the right answer is 210 by trying some programming script. But I don't know what was wrong with my solution. Please help me. Thank you!






combinatorics







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$endgroup$




My solution:



We have:
$x_{1} + x_{2} + x_{3} + x_{4} = 40$ where $2 leq x_{1} leq 8, x_{2} leq 4, x_{3} geq 4, x_{4} leq 5$



$Leftrightarrow x_{2} + x_{3} + x_{4} = 40 - x_{1} quad (*)$



Consider:



$x_{2} + x_{3} + x_{4} = 40 - x_{1}$ where $x_{2} geq 0, x_{3} geq 4, x_{4} geq 0 quad (**)$



$x_{2} + x_{3} + x_{4} = 40 - x_{1}$ where $x_{2} geq 5, x_{3} geq 4, x_{4} geq 6 quad (***)$



Let $f$ is the function that compute the number of non-negative solutions of an equation.



$implies f(*) = f(**) - f(***)$



Thus, the number of non-negative solutions of (*) is $sum_{x_{1} = 2}^{8}( {40 - x_{1} + 3 - 1 choose 3 - 1} - {25-x_{1}+3 - 1 choose 3 - 1}) = 3045$



I found that the right answer is 210 by trying some programming script. But I don't know what was wrong with my solution. Please help me. Thank you!






combinatorics




combinatorics






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asked Dec 9 '18 at 16:06









Duy HuynhDuy Huynh

254




254








  • 1




    $begingroup$
    The three upper limits on $x_1, x_2, x_4$ and that the only variable not upper restrained is dependent on the three restrained makes the stars and bars/choose solution too restrained to be relevent. Do this as pure choose/multiplying. There are so many chooses an the three restrained variables, and the unrestrained can be set up to be completely dependent on the other three. So it is straight mulitplication
    $endgroup$
    – fleablood
    Dec 9 '18 at 16:39






  • 1




    $begingroup$
    YOu are way over counting. You have the solution for $x_2< 5$ OR $x_4< 5$. You don't want that you want AND. ANd you also never take $x_3 ge 4$ into account. The first term should be $36$ not $40$.
    $endgroup$
    – fleablood
    Dec 9 '18 at 16:47












  • $begingroup$
    @fleablood thank for your comment, this is my wrong.
    $endgroup$
    – Duy Huynh
    Dec 9 '18 at 16:51














  • 1




    $begingroup$
    The three upper limits on $x_1, x_2, x_4$ and that the only variable not upper restrained is dependent on the three restrained makes the stars and bars/choose solution too restrained to be relevent. Do this as pure choose/multiplying. There are so many chooses an the three restrained variables, and the unrestrained can be set up to be completely dependent on the other three. So it is straight mulitplication
    $endgroup$
    – fleablood
    Dec 9 '18 at 16:39






  • 1




    $begingroup$
    YOu are way over counting. You have the solution for $x_2< 5$ OR $x_4< 5$. You don't want that you want AND. ANd you also never take $x_3 ge 4$ into account. The first term should be $36$ not $40$.
    $endgroup$
    – fleablood
    Dec 9 '18 at 16:47












  • $begingroup$
    @fleablood thank for your comment, this is my wrong.
    $endgroup$
    – Duy Huynh
    Dec 9 '18 at 16:51








1




1




$begingroup$
The three upper limits on $x_1, x_2, x_4$ and that the only variable not upper restrained is dependent on the three restrained makes the stars and bars/choose solution too restrained to be relevent. Do this as pure choose/multiplying. There are so many chooses an the three restrained variables, and the unrestrained can be set up to be completely dependent on the other three. So it is straight mulitplication
$endgroup$
– fleablood
Dec 9 '18 at 16:39




$begingroup$
The three upper limits on $x_1, x_2, x_4$ and that the only variable not upper restrained is dependent on the three restrained makes the stars and bars/choose solution too restrained to be relevent. Do this as pure choose/multiplying. There are so many chooses an the three restrained variables, and the unrestrained can be set up to be completely dependent on the other three. So it is straight mulitplication
$endgroup$
– fleablood
Dec 9 '18 at 16:39




1




1




$begingroup$
YOu are way over counting. You have the solution for $x_2< 5$ OR $x_4< 5$. You don't want that you want AND. ANd you also never take $x_3 ge 4$ into account. The first term should be $36$ not $40$.
$endgroup$
– fleablood
Dec 9 '18 at 16:47






$begingroup$
YOu are way over counting. You have the solution for $x_2< 5$ OR $x_4< 5$. You don't want that you want AND. ANd you also never take $x_3 ge 4$ into account. The first term should be $36$ not $40$.
$endgroup$
– fleablood
Dec 9 '18 at 16:47














$begingroup$
@fleablood thank for your comment, this is my wrong.
$endgroup$
– Duy Huynh
Dec 9 '18 at 16:51




$begingroup$
@fleablood thank for your comment, this is my wrong.
$endgroup$
– Duy Huynh
Dec 9 '18 at 16:51










4 Answers
4






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4












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$x_1 + x_2 + x_4 le 8 + 4 + 5 = 17$ so $x_3=40 - x_1 + s_2 + x_4 ge 40 -17 ge 4$ so we can ignore the restriction on $x_3$.



$2 le x_1 le 6$ so there are $7$ values that $x_1$ can be, $xle 4$ so there are $5$ values it can be. $x_4 le 5$ so there are $6$ values it can be and $x_3$ must be $40 - x_1 - x_2 -x_4$ there is only one option dependant on the other three options.



So there $7*5*6 = 210$ options.



Your solution dosn't take $x_3 ge 4$ into account (which you can be seting it up so that the some is $36$ and not $40$-- I haven't done the math to figure it out but that will lower you answer significantly. Also by subtracting you are removing the cases with both $x_2 ge 5$ and $x_4 ge 6$ but not removing the cases where one or the other is.)



I think to fix your problem using inclusion exclusion you'd want



$sum_{x_1=2}^8({{40 - 4 -x_1 + 3-1}choose {3-1}} - {{40 - 4-5 -x_1 + 3-1}choose {3-1}}-{{40 - 4-6 -x_1 + 3-1}choose {3-1}}+{{40 - 4 -5-6-x_1 + 3-1}choose {3-1}})=$



And I'm too lazy to finish.






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    newcommand{verts}[1]{leftvert,{#1},rightvert}$
    A General Method:
    begin{align}
    &bbox[10px,#ffd]{sum_{x_{1} = 2}^{8}
    sum_{x_{2} = 0}^{4}sum_{x_{3} = 4}^{infty}sum_{x_{4} = 0}^{5}
    bracks{z^{40}}z^{x_{1} + x_{2} + x_{3} + x_{4}}} =
    sum_{x_{1} = 0}^{6}
    sum_{x_{2} = 0}^{4}sum_{x_{3} = 0}^{infty}sum_{x_{4} = 0}^{5}
    bracks{z^{30}}z^{x_{1} + x_{2} + x_{3} + x_{4}}
    \[5mm] = &
    bracks{z^{30}}pars{sum_{x_{1} = 0}^{6}z^{x_{1}}}
    pars{sum_{x_{2} = 0}^{4}z^{x_{2}}}
    pars{sum_{x_{3} = 0}^{infty}z^{x_{3}}}
    pars{sum_{x_{1} = 0}^{5}z^{x_{4}}}
    \[5mm] = &
    bracks{z^{40}}{1 - z^{7} over 1 - z},{1 - z^{5} over 1 - z},{1 over 1 - z},{1 - z^{6} over 1 - z}
    \[5mm] = &
    bracks{z^{40}}pars{-z^{18} + z^{13} + z^{12} + z^{11} - z^{7} - z^{6} - z^{5} + 1}pars{1 - z}^{-4}
    \[5mm] = &
    -{-4 choose 22} - {-4 choose 27} + {-4 choose 28} -
    {-4 choose 29} + {-4 choose 33} - {-4 choose 34} +
    {-4 choose 35} + {-4 choose 40}
    \[5mm] = &
    - underbrace{25 choose 22}_{ds{2300}} +
    underbrace{30 choose 27}_{ds{4060}} +
    underbrace{31 choose 28}_{ds{4495}} +
    underbrace{32 choose 29}_{ds{4960}} -
    underbrace{36 choose 33}_{ds{7140}} -
    underbrace{37 choose 34}_{ds{7770}} -
    \[2mm] & -
    underbrace{38 choose 35}_{ds{8436}} +
    underbrace{43 choose 40}_{ds{12341}} = bbx{large 210}
    end{align}






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      3












      $begingroup$

      It's the coefficient of $x^{40}$ of the product polynomial



      $$(x^2+x^3+x^4 +x^5 + x^6 + x^7 + x^8)(1+x^1+x^2+x^3 +x^4)(x^4 + x^5 + ldots)(1+x^1+x^2+x^3 +x^4 + x^5)$$



      Or equivalently the coefficient of $x^{34}$ of



      $$(1+x+x^2+x^3+x^4 +x^5 + x^6 )(1+x^1+x^2+x^3 +x^4)(1 + x + x^2 + ldots)(1+x^1+x^2+x^3 +x^4 + x^5)$$



      which can be found using (generalised) binomials etc.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Isn't this just rephrasing the problem?
        $endgroup$
        – Christoph
        Dec 9 '18 at 16:56










      • $begingroup$
        @Christoph just giving an alternative path.
        $endgroup$
        – Henno Brandsma
        Dec 9 '18 at 16:58



















      2












      $begingroup$

      Math answer



      Note that the given constraints for $x_1, x_2$ and $x_4$ and $sumlimits_{i = 1}^4 x_i = 40$ allows us to define
      $$
      begin{aligned}
      x_3 &= 40 - x_1 - x_2 - x_4 \
      &ge 40 - 8 - 4 - 5 \
      &= 23.
      end{aligned}$$
      This renders the constraint $x_3 ge 4$ redundant. As a result, the required answer is $(8-2+1) times (4+1) times (5+1) = 210$.




      Julia Programming Script



      x1 = 2:8
      
x2 = 0:4
      
x4 = 0:5
      
x3 = [40 - i - j - k for i in x1 for j in x2 for k in x4]
      
println(minimum(x3))  # returns 23
      
println(length(x3))   # returns 210


      Test this script on Tutorial's Point's online compiler.






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        4 Answers
        4






        active

        oldest

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        4 Answers
        4






        active

        oldest

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        active

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        active

        oldest

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        4












        $begingroup$

        $x_1 + x_2 + x_4 le 8 + 4 + 5 = 17$ so $x_3=40 - x_1 + s_2 + x_4 ge 40 -17 ge 4$ so we can ignore the restriction on $x_3$.



        $2 le x_1 le 6$ so there are $7$ values that $x_1$ can be, $xle 4$ so there are $5$ values it can be. $x_4 le 5$ so there are $6$ values it can be and $x_3$ must be $40 - x_1 - x_2 -x_4$ there is only one option dependant on the other three options.



        So there $7*5*6 = 210$ options.



        Your solution dosn't take $x_3 ge 4$ into account (which you can be seting it up so that the some is $36$ and not $40$-- I haven't done the math to figure it out but that will lower you answer significantly. Also by subtracting you are removing the cases with both $x_2 ge 5$ and $x_4 ge 6$ but not removing the cases where one or the other is.)



        I think to fix your problem using inclusion exclusion you'd want



        $sum_{x_1=2}^8({{40 - 4 -x_1 + 3-1}choose {3-1}} - {{40 - 4-5 -x_1 + 3-1}choose {3-1}}-{{40 - 4-6 -x_1 + 3-1}choose {3-1}}+{{40 - 4 -5-6-x_1 + 3-1}choose {3-1}})=$



        And I'm too lazy to finish.






        share|cite|improve this answer











        $endgroup$


















          4












          $begingroup$

          $x_1 + x_2 + x_4 le 8 + 4 + 5 = 17$ so $x_3=40 - x_1 + s_2 + x_4 ge 40 -17 ge 4$ so we can ignore the restriction on $x_3$.



          $2 le x_1 le 6$ so there are $7$ values that $x_1$ can be, $xle 4$ so there are $5$ values it can be. $x_4 le 5$ so there are $6$ values it can be and $x_3$ must be $40 - x_1 - x_2 -x_4$ there is only one option dependant on the other three options.



          So there $7*5*6 = 210$ options.



          Your solution dosn't take $x_3 ge 4$ into account (which you can be seting it up so that the some is $36$ and not $40$-- I haven't done the math to figure it out but that will lower you answer significantly. Also by subtracting you are removing the cases with both $x_2 ge 5$ and $x_4 ge 6$ but not removing the cases where one or the other is.)



          I think to fix your problem using inclusion exclusion you'd want



          $sum_{x_1=2}^8({{40 - 4 -x_1 + 3-1}choose {3-1}} - {{40 - 4-5 -x_1 + 3-1}choose {3-1}}-{{40 - 4-6 -x_1 + 3-1}choose {3-1}}+{{40 - 4 -5-6-x_1 + 3-1}choose {3-1}})=$



          And I'm too lazy to finish.






          share|cite|improve this answer











          $endgroup$
















            4












            4








            4





            $begingroup$

            $x_1 + x_2 + x_4 le 8 + 4 + 5 = 17$ so $x_3=40 - x_1 + s_2 + x_4 ge 40 -17 ge 4$ so we can ignore the restriction on $x_3$.



            $2 le x_1 le 6$ so there are $7$ values that $x_1$ can be, $xle 4$ so there are $5$ values it can be. $x_4 le 5$ so there are $6$ values it can be and $x_3$ must be $40 - x_1 - x_2 -x_4$ there is only one option dependant on the other three options.



            So there $7*5*6 = 210$ options.



            Your solution dosn't take $x_3 ge 4$ into account (which you can be seting it up so that the some is $36$ and not $40$-- I haven't done the math to figure it out but that will lower you answer significantly. Also by subtracting you are removing the cases with both $x_2 ge 5$ and $x_4 ge 6$ but not removing the cases where one or the other is.)



            I think to fix your problem using inclusion exclusion you'd want



            $sum_{x_1=2}^8({{40 - 4 -x_1 + 3-1}choose {3-1}} - {{40 - 4-5 -x_1 + 3-1}choose {3-1}}-{{40 - 4-6 -x_1 + 3-1}choose {3-1}}+{{40 - 4 -5-6-x_1 + 3-1}choose {3-1}})=$



            And I'm too lazy to finish.






            share|cite|improve this answer











            $endgroup$



            $x_1 + x_2 + x_4 le 8 + 4 + 5 = 17$ so $x_3=40 - x_1 + s_2 + x_4 ge 40 -17 ge 4$ so we can ignore the restriction on $x_3$.



            $2 le x_1 le 6$ so there are $7$ values that $x_1$ can be, $xle 4$ so there are $5$ values it can be. $x_4 le 5$ so there are $6$ values it can be and $x_3$ must be $40 - x_1 - x_2 -x_4$ there is only one option dependant on the other three options.



            So there $7*5*6 = 210$ options.



            Your solution dosn't take $x_3 ge 4$ into account (which you can be seting it up so that the some is $36$ and not $40$-- I haven't done the math to figure it out but that will lower you answer significantly. Also by subtracting you are removing the cases with both $x_2 ge 5$ and $x_4 ge 6$ but not removing the cases where one or the other is.)



            I think to fix your problem using inclusion exclusion you'd want



            $sum_{x_1=2}^8({{40 - 4 -x_1 + 3-1}choose {3-1}} - {{40 - 4-5 -x_1 + 3-1}choose {3-1}}-{{40 - 4-6 -x_1 + 3-1}choose {3-1}}+{{40 - 4 -5-6-x_1 + 3-1}choose {3-1}})=$



            And I'm too lazy to finish.







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            edited Dec 9 '18 at 17:08

























            answered Dec 9 '18 at 16:32









            fleabloodfleablood

            71k22686




            71k22686























                4












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                newcommand{pars}[1]{left(,{#1},right)}
                newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
                newcommand{root}[2]{,sqrt[#1]{,{#2},},}
                newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
                newcommand{verts}[1]{leftvert,{#1},rightvert}$
                A General Method:
                begin{align}
                &bbox[10px,#ffd]{sum_{x_{1} = 2}^{8}
                sum_{x_{2} = 0}^{4}sum_{x_{3} = 4}^{infty}sum_{x_{4} = 0}^{5}
                bracks{z^{40}}z^{x_{1} + x_{2} + x_{3} + x_{4}}} =
                sum_{x_{1} = 0}^{6}
                sum_{x_{2} = 0}^{4}sum_{x_{3} = 0}^{infty}sum_{x_{4} = 0}^{5}
                bracks{z^{30}}z^{x_{1} + x_{2} + x_{3} + x_{4}}
                \[5mm] = &
                bracks{z^{30}}pars{sum_{x_{1} = 0}^{6}z^{x_{1}}}
                pars{sum_{x_{2} = 0}^{4}z^{x_{2}}}
                pars{sum_{x_{3} = 0}^{infty}z^{x_{3}}}
                pars{sum_{x_{1} = 0}^{5}z^{x_{4}}}
                \[5mm] = &
                bracks{z^{40}}{1 - z^{7} over 1 - z},{1 - z^{5} over 1 - z},{1 over 1 - z},{1 - z^{6} over 1 - z}
                \[5mm] = &
                bracks{z^{40}}pars{-z^{18} + z^{13} + z^{12} + z^{11} - z^{7} - z^{6} - z^{5} + 1}pars{1 - z}^{-4}
                \[5mm] = &
                -{-4 choose 22} - {-4 choose 27} + {-4 choose 28} -
                {-4 choose 29} + {-4 choose 33} - {-4 choose 34} +
                {-4 choose 35} + {-4 choose 40}
                \[5mm] = &
                - underbrace{25 choose 22}_{ds{2300}} +
                underbrace{30 choose 27}_{ds{4060}} +
                underbrace{31 choose 28}_{ds{4495}} +
                underbrace{32 choose 29}_{ds{4960}} -
                underbrace{36 choose 33}_{ds{7140}} -
                underbrace{37 choose 34}_{ds{7770}} -
                \[2mm] & -
                underbrace{38 choose 35}_{ds{8436}} +
                underbrace{43 choose 40}_{ds{12341}} = bbx{large 210}
                end{align}






                share|cite|improve this answer









                $endgroup$


















                  4












                  $begingroup$

                  $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
                  newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
                  newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
                  newcommand{dd}{mathrm{d}}
                  newcommand{ds}[1]{displaystyle{#1}}
                  newcommand{expo}[1]{,mathrm{e}^{#1},}
                  newcommand{ic}{mathrm{i}}
                  newcommand{mc}[1]{mathcal{#1}}
                  newcommand{mrm}[1]{mathrm{#1}}
                  newcommand{pars}[1]{left(,{#1},right)}
                  newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
                  newcommand{root}[2]{,sqrt[#1]{,{#2},},}
                  newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
                  newcommand{verts}[1]{leftvert,{#1},rightvert}$
                  A General Method:
                  begin{align}
                  &bbox[10px,#ffd]{sum_{x_{1} = 2}^{8}
                  sum_{x_{2} = 0}^{4}sum_{x_{3} = 4}^{infty}sum_{x_{4} = 0}^{5}
                  bracks{z^{40}}z^{x_{1} + x_{2} + x_{3} + x_{4}}} =
                  sum_{x_{1} = 0}^{6}
                  sum_{x_{2} = 0}^{4}sum_{x_{3} = 0}^{infty}sum_{x_{4} = 0}^{5}
                  bracks{z^{30}}z^{x_{1} + x_{2} + x_{3} + x_{4}}
                  \[5mm] = &
                  bracks{z^{30}}pars{sum_{x_{1} = 0}^{6}z^{x_{1}}}
                  pars{sum_{x_{2} = 0}^{4}z^{x_{2}}}
                  pars{sum_{x_{3} = 0}^{infty}z^{x_{3}}}
                  pars{sum_{x_{1} = 0}^{5}z^{x_{4}}}
                  \[5mm] = &
                  bracks{z^{40}}{1 - z^{7} over 1 - z},{1 - z^{5} over 1 - z},{1 over 1 - z},{1 - z^{6} over 1 - z}
                  \[5mm] = &
                  bracks{z^{40}}pars{-z^{18} + z^{13} + z^{12} + z^{11} - z^{7} - z^{6} - z^{5} + 1}pars{1 - z}^{-4}
                  \[5mm] = &
                  -{-4 choose 22} - {-4 choose 27} + {-4 choose 28} -
                  {-4 choose 29} + {-4 choose 33} - {-4 choose 34} +
                  {-4 choose 35} + {-4 choose 40}
                  \[5mm] = &
                  - underbrace{25 choose 22}_{ds{2300}} +
                  underbrace{30 choose 27}_{ds{4060}} +
                  underbrace{31 choose 28}_{ds{4495}} +
                  underbrace{32 choose 29}_{ds{4960}} -
                  underbrace{36 choose 33}_{ds{7140}} -
                  underbrace{37 choose 34}_{ds{7770}} -
                  \[2mm] & -
                  underbrace{38 choose 35}_{ds{8436}} +
                  underbrace{43 choose 40}_{ds{12341}} = bbx{large 210}
                  end{align}






                  share|cite|improve this answer









                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
                    newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
                    newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
                    newcommand{dd}{mathrm{d}}
                    newcommand{ds}[1]{displaystyle{#1}}
                    newcommand{expo}[1]{,mathrm{e}^{#1},}
                    newcommand{ic}{mathrm{i}}
                    newcommand{mc}[1]{mathcal{#1}}
                    newcommand{mrm}[1]{mathrm{#1}}
                    newcommand{pars}[1]{left(,{#1},right)}
                    newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
                    newcommand{root}[2]{,sqrt[#1]{,{#2},},}
                    newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
                    newcommand{verts}[1]{leftvert,{#1},rightvert}$
                    A General Method:
                    begin{align}
                    &bbox[10px,#ffd]{sum_{x_{1} = 2}^{8}
                    sum_{x_{2} = 0}^{4}sum_{x_{3} = 4}^{infty}sum_{x_{4} = 0}^{5}
                    bracks{z^{40}}z^{x_{1} + x_{2} + x_{3} + x_{4}}} =
                    sum_{x_{1} = 0}^{6}
                    sum_{x_{2} = 0}^{4}sum_{x_{3} = 0}^{infty}sum_{x_{4} = 0}^{5}
                    bracks{z^{30}}z^{x_{1} + x_{2} + x_{3} + x_{4}}
                    \[5mm] = &
                    bracks{z^{30}}pars{sum_{x_{1} = 0}^{6}z^{x_{1}}}
                    pars{sum_{x_{2} = 0}^{4}z^{x_{2}}}
                    pars{sum_{x_{3} = 0}^{infty}z^{x_{3}}}
                    pars{sum_{x_{1} = 0}^{5}z^{x_{4}}}
                    \[5mm] = &
                    bracks{z^{40}}{1 - z^{7} over 1 - z},{1 - z^{5} over 1 - z},{1 over 1 - z},{1 - z^{6} over 1 - z}
                    \[5mm] = &
                    bracks{z^{40}}pars{-z^{18} + z^{13} + z^{12} + z^{11} - z^{7} - z^{6} - z^{5} + 1}pars{1 - z}^{-4}
                    \[5mm] = &
                    -{-4 choose 22} - {-4 choose 27} + {-4 choose 28} -
                    {-4 choose 29} + {-4 choose 33} - {-4 choose 34} +
                    {-4 choose 35} + {-4 choose 40}
                    \[5mm] = &
                    - underbrace{25 choose 22}_{ds{2300}} +
                    underbrace{30 choose 27}_{ds{4060}} +
                    underbrace{31 choose 28}_{ds{4495}} +
                    underbrace{32 choose 29}_{ds{4960}} -
                    underbrace{36 choose 33}_{ds{7140}} -
                    underbrace{37 choose 34}_{ds{7770}} -
                    \[2mm] & -
                    underbrace{38 choose 35}_{ds{8436}} +
                    underbrace{43 choose 40}_{ds{12341}} = bbx{large 210}
                    end{align}






                    share|cite|improve this answer









                    $endgroup$



                    $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
                    newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
                    newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
                    newcommand{dd}{mathrm{d}}
                    newcommand{ds}[1]{displaystyle{#1}}
                    newcommand{expo}[1]{,mathrm{e}^{#1},}
                    newcommand{ic}{mathrm{i}}
                    newcommand{mc}[1]{mathcal{#1}}
                    newcommand{mrm}[1]{mathrm{#1}}
                    newcommand{pars}[1]{left(,{#1},right)}
                    newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
                    newcommand{root}[2]{,sqrt[#1]{,{#2},},}
                    newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
                    newcommand{verts}[1]{leftvert,{#1},rightvert}$
                    A General Method:
                    begin{align}
                    &bbox[10px,#ffd]{sum_{x_{1} = 2}^{8}
                    sum_{x_{2} = 0}^{4}sum_{x_{3} = 4}^{infty}sum_{x_{4} = 0}^{5}
                    bracks{z^{40}}z^{x_{1} + x_{2} + x_{3} + x_{4}}} =
                    sum_{x_{1} = 0}^{6}
                    sum_{x_{2} = 0}^{4}sum_{x_{3} = 0}^{infty}sum_{x_{4} = 0}^{5}
                    bracks{z^{30}}z^{x_{1} + x_{2} + x_{3} + x_{4}}
                    \[5mm] = &
                    bracks{z^{30}}pars{sum_{x_{1} = 0}^{6}z^{x_{1}}}
                    pars{sum_{x_{2} = 0}^{4}z^{x_{2}}}
                    pars{sum_{x_{3} = 0}^{infty}z^{x_{3}}}
                    pars{sum_{x_{1} = 0}^{5}z^{x_{4}}}
                    \[5mm] = &
                    bracks{z^{40}}{1 - z^{7} over 1 - z},{1 - z^{5} over 1 - z},{1 over 1 - z},{1 - z^{6} over 1 - z}
                    \[5mm] = &
                    bracks{z^{40}}pars{-z^{18} + z^{13} + z^{12} + z^{11} - z^{7} - z^{6} - z^{5} + 1}pars{1 - z}^{-4}
                    \[5mm] = &
                    -{-4 choose 22} - {-4 choose 27} + {-4 choose 28} -
                    {-4 choose 29} + {-4 choose 33} - {-4 choose 34} +
                    {-4 choose 35} + {-4 choose 40}
                    \[5mm] = &
                    - underbrace{25 choose 22}_{ds{2300}} +
                    underbrace{30 choose 27}_{ds{4060}} +
                    underbrace{31 choose 28}_{ds{4495}} +
                    underbrace{32 choose 29}_{ds{4960}} -
                    underbrace{36 choose 33}_{ds{7140}} -
                    underbrace{37 choose 34}_{ds{7770}} -
                    \[2mm] & -
                    underbrace{38 choose 35}_{ds{8436}} +
                    underbrace{43 choose 40}_{ds{12341}} = bbx{large 210}
                    end{align}







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 9 '18 at 17:54









                    Felix MarinFelix Marin

                    68k7108142




                    68k7108142























                        3












                        $begingroup$

                        It's the coefficient of $x^{40}$ of the product polynomial



                        $$(x^2+x^3+x^4 +x^5 + x^6 + x^7 + x^8)(1+x^1+x^2+x^3 +x^4)(x^4 + x^5 + ldots)(1+x^1+x^2+x^3 +x^4 + x^5)$$



                        Or equivalently the coefficient of $x^{34}$ of



                        $$(1+x+x^2+x^3+x^4 +x^5 + x^6 )(1+x^1+x^2+x^3 +x^4)(1 + x + x^2 + ldots)(1+x^1+x^2+x^3 +x^4 + x^5)$$



                        which can be found using (generalised) binomials etc.






                        share|cite|improve this answer









                        $endgroup$













                        • $begingroup$
                          Isn't this just rephrasing the problem?
                          $endgroup$
                          – Christoph
                          Dec 9 '18 at 16:56










                        • $begingroup$
                          @Christoph just giving an alternative path.
                          $endgroup$
                          – Henno Brandsma
                          Dec 9 '18 at 16:58
















                        3












                        $begingroup$

                        It's the coefficient of $x^{40}$ of the product polynomial



                        $$(x^2+x^3+x^4 +x^5 + x^6 + x^7 + x^8)(1+x^1+x^2+x^3 +x^4)(x^4 + x^5 + ldots)(1+x^1+x^2+x^3 +x^4 + x^5)$$



                        Or equivalently the coefficient of $x^{34}$ of



                        $$(1+x+x^2+x^3+x^4 +x^5 + x^6 )(1+x^1+x^2+x^3 +x^4)(1 + x + x^2 + ldots)(1+x^1+x^2+x^3 +x^4 + x^5)$$



                        which can be found using (generalised) binomials etc.






                        share|cite|improve this answer









                        $endgroup$













                        • $begingroup$
                          Isn't this just rephrasing the problem?
                          $endgroup$
                          – Christoph
                          Dec 9 '18 at 16:56










                        • $begingroup$
                          @Christoph just giving an alternative path.
                          $endgroup$
                          – Henno Brandsma
                          Dec 9 '18 at 16:58














                        3












                        3








                        3





                        $begingroup$

                        It's the coefficient of $x^{40}$ of the product polynomial



                        $$(x^2+x^3+x^4 +x^5 + x^6 + x^7 + x^8)(1+x^1+x^2+x^3 +x^4)(x^4 + x^5 + ldots)(1+x^1+x^2+x^3 +x^4 + x^5)$$



                        Or equivalently the coefficient of $x^{34}$ of



                        $$(1+x+x^2+x^3+x^4 +x^5 + x^6 )(1+x^1+x^2+x^3 +x^4)(1 + x + x^2 + ldots)(1+x^1+x^2+x^3 +x^4 + x^5)$$



                        which can be found using (generalised) binomials etc.






                        share|cite|improve this answer









                        $endgroup$



                        It's the coefficient of $x^{40}$ of the product polynomial



                        $$(x^2+x^3+x^4 +x^5 + x^6 + x^7 + x^8)(1+x^1+x^2+x^3 +x^4)(x^4 + x^5 + ldots)(1+x^1+x^2+x^3 +x^4 + x^5)$$



                        Or equivalently the coefficient of $x^{34}$ of



                        $$(1+x+x^2+x^3+x^4 +x^5 + x^6 )(1+x^1+x^2+x^3 +x^4)(1 + x + x^2 + ldots)(1+x^1+x^2+x^3 +x^4 + x^5)$$



                        which can be found using (generalised) binomials etc.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Dec 9 '18 at 16:12









                        Henno BrandsmaHenno Brandsma

                        110k347116




                        110k347116












                        • $begingroup$
                          Isn't this just rephrasing the problem?
                          $endgroup$
                          – Christoph
                          Dec 9 '18 at 16:56










                        • $begingroup$
                          @Christoph just giving an alternative path.
                          $endgroup$
                          – Henno Brandsma
                          Dec 9 '18 at 16:58


















                        • $begingroup$
                          Isn't this just rephrasing the problem?
                          $endgroup$
                          – Christoph
                          Dec 9 '18 at 16:56










                        • $begingroup$
                          @Christoph just giving an alternative path.
                          $endgroup$
                          – Henno Brandsma
                          Dec 9 '18 at 16:58
















                        $begingroup$
                        Isn't this just rephrasing the problem?
                        $endgroup$
                        – Christoph
                        Dec 9 '18 at 16:56




                        $begingroup$
                        Isn't this just rephrasing the problem?
                        $endgroup$
                        – Christoph
                        Dec 9 '18 at 16:56












                        $begingroup$
                        @Christoph just giving an alternative path.
                        $endgroup$
                        – Henno Brandsma
                        Dec 9 '18 at 16:58




                        $begingroup$
                        @Christoph just giving an alternative path.
                        $endgroup$
                        – Henno Brandsma
                        Dec 9 '18 at 16:58











                        2












                        $begingroup$

                        Math answer



                        Note that the given constraints for $x_1, x_2$ and $x_4$ and $sumlimits_{i = 1}^4 x_i = 40$ allows us to define
                        $$
                        begin{aligned}
                        x_3 &= 40 - x_1 - x_2 - x_4 \
                        &ge 40 - 8 - 4 - 5 \
                        &= 23.
                        end{aligned}$$
                        This renders the constraint $x_3 ge 4$ redundant. As a result, the required answer is $(8-2+1) times (4+1) times (5+1) = 210$.




                        Julia Programming Script



                        x1 = 2:8
                        
x2 = 0:4
                        
x4 = 0:5
                        
x3 = [40 - i - j - k for i in x1 for j in x2 for k in x4]
                        
println(minimum(x3))  # returns 23
                        
println(length(x3))   # returns 210


                        Test this script on Tutorial's Point's online compiler.






                        share|cite|improve this answer











                        $endgroup$


















                          2












                          $begingroup$

                          Math answer



                          Note that the given constraints for $x_1, x_2$ and $x_4$ and $sumlimits_{i = 1}^4 x_i = 40$ allows us to define
                          $$
                          begin{aligned}
                          x_3 &= 40 - x_1 - x_2 - x_4 \
                          &ge 40 - 8 - 4 - 5 \
                          &= 23.
                          end{aligned}$$
                          This renders the constraint $x_3 ge 4$ redundant. As a result, the required answer is $(8-2+1) times (4+1) times (5+1) = 210$.




                          Julia Programming Script



                          x1 = 2:8
                          
x2 = 0:4
                          
x4 = 0:5
                          
x3 = [40 - i - j - k for i in x1 for j in x2 for k in x4]
                          
println(minimum(x3))  # returns 23
                          
println(length(x3))   # returns 210


                          Test this script on Tutorial's Point's online compiler.






                          share|cite|improve this answer











                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            Math answer



                            Note that the given constraints for $x_1, x_2$ and $x_4$ and $sumlimits_{i = 1}^4 x_i = 40$ allows us to define
                            $$
                            begin{aligned}
                            x_3 &= 40 - x_1 - x_2 - x_4 \
                            &ge 40 - 8 - 4 - 5 \
                            &= 23.
                            end{aligned}$$
                            This renders the constraint $x_3 ge 4$ redundant. As a result, the required answer is $(8-2+1) times (4+1) times (5+1) = 210$.




                            Julia Programming Script



                            x1 = 2:8
                            
x2 = 0:4
                            
x4 = 0:5
                            
x3 = [40 - i - j - k for i in x1 for j in x2 for k in x4]
                            
println(minimum(x3))  # returns 23
                            
println(length(x3))   # returns 210


                            Test this script on Tutorial's Point's online compiler.






                            share|cite|improve this answer











                            $endgroup$



                            Math answer



                            Note that the given constraints for $x_1, x_2$ and $x_4$ and $sumlimits_{i = 1}^4 x_i = 40$ allows us to define
                            $$
                            begin{aligned}
                            x_3 &= 40 - x_1 - x_2 - x_4 \
                            &ge 40 - 8 - 4 - 5 \
                            &= 23.
                            end{aligned}$$
                            This renders the constraint $x_3 ge 4$ redundant. As a result, the required answer is $(8-2+1) times (4+1) times (5+1) = 210$.




                            Julia Programming Script



                            x1 = 2:8
                            
x2 = 0:4
                            
x4 = 0:5
                            
x3 = [40 - i - j - k for i in x1 for j in x2 for k in x4]
                            
println(minimum(x3))  # returns 23
                            
println(length(x3))   # returns 210


                            Test this script on Tutorial's Point's online compiler.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Dec 9 '18 at 16:40

























                            answered Dec 9 '18 at 16:32









                            GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會

                            13.3k72549




                            13.3k72549






























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