counting ways for sum with maximums on variables












0














If $a < 3$ and $b < 4$ and $a, b, c, d geq 0$, how may different combinations are there for $a + b + c + d = 10$?



I've tried star and bar method but had no luck. How can I approach this?






combinatorics combinations







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  • Welcome to MathSE. When you pose a question here, you should show what you have attempted and explain where you are stuck. In this case, you could show your calculation for the number of nonnegative integer solutions of the equation $a + b + c + d = 10$ so that we can check that calculation. I suspect that you are stuck on how to handle the restrictions, but you should explicitly state that so that you receive responses that address the specific difficulties you are encountering.
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0














If $a < 3$ and $b < 4$ and $a, b, c, d geq 0$, how may different combinations are there for $a + b + c + d = 10$?



I've tried star and bar method but had no luck. How can I approach this?






combinatorics combinations







share|cite|improve this question
























  • Welcome to MathSE. When you pose a question here, you should show what you have attempted and explain where you are stuck. In this case, you could show your calculation for the number of nonnegative integer solutions of the equation $a + b + c + d = 10$ so that we can check that calculation. I suspect that you are stuck on how to handle the restrictions, but you should explicitly state that so that you receive responses that address the specific difficulties you are encountering.
    – N. F. Taussig
    Nov 25 at 11:27










  • This tutorial explains how to typeset mathematics on this site.
    – N. F. Taussig
    Nov 25 at 11:28














0












0








0


1





If $a < 3$ and $b < 4$ and $a, b, c, d geq 0$, how may different combinations are there for $a + b + c + d = 10$?



I've tried star and bar method but had no luck. How can I approach this?






combinatorics combinations







share|cite|improve this question















If $a < 3$ and $b < 4$ and $a, b, c, d geq 0$, how may different combinations are there for $a + b + c + d = 10$?



I've tried star and bar method but had no luck. How can I approach this?






combinatorics combinations




combinatorics combinations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 at 11:29









N. F. Taussig

43.5k93355




43.5k93355










asked Nov 24 at 19:29









WEPIOD

31




31












  • Welcome to MathSE. When you pose a question here, you should show what you have attempted and explain where you are stuck. In this case, you could show your calculation for the number of nonnegative integer solutions of the equation $a + b + c + d = 10$ so that we can check that calculation. I suspect that you are stuck on how to handle the restrictions, but you should explicitly state that so that you receive responses that address the specific difficulties you are encountering.
    – N. F. Taussig
    Nov 25 at 11:27










  • This tutorial explains how to typeset mathematics on this site.
    – N. F. Taussig
    Nov 25 at 11:28


















  • Welcome to MathSE. When you pose a question here, you should show what you have attempted and explain where you are stuck. In this case, you could show your calculation for the number of nonnegative integer solutions of the equation $a + b + c + d = 10$ so that we can check that calculation. I suspect that you are stuck on how to handle the restrictions, but you should explicitly state that so that you receive responses that address the specific difficulties you are encountering.
    – N. F. Taussig
    Nov 25 at 11:27










  • This tutorial explains how to typeset mathematics on this site.
    – N. F. Taussig
    Nov 25 at 11:28
















Welcome to MathSE. When you pose a question here, you should show what you have attempted and explain where you are stuck. In this case, you could show your calculation for the number of nonnegative integer solutions of the equation $a + b + c + d = 10$ so that we can check that calculation. I suspect that you are stuck on how to handle the restrictions, but you should explicitly state that so that you receive responses that address the specific difficulties you are encountering.
– N. F. Taussig
Nov 25 at 11:27




Welcome to MathSE. When you pose a question here, you should show what you have attempted and explain where you are stuck. In this case, you could show your calculation for the number of nonnegative integer solutions of the equation $a + b + c + d = 10$ so that we can check that calculation. I suspect that you are stuck on how to handle the restrictions, but you should explicitly state that so that you receive responses that address the specific difficulties you are encountering.
– N. F. Taussig
Nov 25 at 11:27












This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Nov 25 at 11:28




This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Nov 25 at 11:28










2 Answers
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I am assuming $a,b,c,d in mathbb{Z}$ and non-negative. And even though you mentioned combination, I assume $a=b=c=0,d=10$ is different from $a=b=d=0, c=10$. It very easy to tune it if you intended otherwise.



This problem can be most elegantly solved with additive number theory method. As I don't know it (yet :), I will approach it with a computer algorithm method called dynamic programming.



Define $p(n,m)$ the number of ways of partitioning an integer $n$ into sum of $m$ non-negative integers. $p(n,m)$ can be arranged into a $ntimes m$ matrix/grid to minimize counting work.



So our problem becomes evaluation of $p(10,4)$ with the restriction $a<3, b<4$.



$p(10,4)=p(10,3)+p(9,3)+p(8,3)$ RHS, assign $0,1,2$ to $a$ respectively and pass to next level



$p(10,3)=p(10,2)+p(9,2)+p(8,2)+p(7,2)$ RHS, assign $0,1,2,3$ to $b$ respectively and pass to next level；



$p(9,3)=p(9,2)+p(8,2)+p(7,2)+p(6,2)$ RHS, assign $0,1,2,3$ to $b$ respectively and pass to next level;



(Note: $p(9,2),p(8,2)$ etc appears on RHS multiple times, the intention of a dynamic programming is to process these repetition only once.)



$cdots$



$p(10,2)=p(10,1)+p(9,1)+cdots+p(0,1)=11$, assign $0,1,2,cdots,10$ to $c$ respectively. And each represent a final scheme, so we have $11$ ways...



Make a table and do it manually. It's a good educational experience to understand dynamic programming.






share|cite|improve this answer































    0














    There are three possible values $(0,1,2)$ for $a$ and four $(0,1,2,3)$ for $b$. Thus there are twelve settings for $a$ and $b$. For each setting, there are $10 - (a+b) + 1$ possible values for $c$. Then $d$ is constrained and fixed.



    So go through the $12$ settings for $a$ and $b$, and for each, add up the number of settings for $c$. And you're done!



    For instance: $a=0, b=0$ leads to $11$ possible values for $c$ $(0, 1, ldots, 10)$ and then $d$ is constrained.



    I get $102$.






    share|cite|improve this answer























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      2 Answers
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      2 Answers
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      1














      I am assuming $a,b,c,d in mathbb{Z}$ and non-negative. And even though you mentioned combination, I assume $a=b=c=0,d=10$ is different from $a=b=d=0, c=10$. It very easy to tune it if you intended otherwise.



      This problem can be most elegantly solved with additive number theory method. As I don't know it (yet :), I will approach it with a computer algorithm method called dynamic programming.



      Define $p(n,m)$ the number of ways of partitioning an integer $n$ into sum of $m$ non-negative integers. $p(n,m)$ can be arranged into a $ntimes m$ matrix/grid to minimize counting work.



      So our problem becomes evaluation of $p(10,4)$ with the restriction $a<3, b<4$.



      $p(10,4)=p(10,3)+p(9,3)+p(8,3)$ RHS, assign $0,1,2$ to $a$ respectively and pass to next level



      $p(10,3)=p(10,2)+p(9,2)+p(8,2)+p(7,2)$ RHS, assign $0,1,2,3$ to $b$ respectively and pass to next level；



      $p(9,3)=p(9,2)+p(8,2)+p(7,2)+p(6,2)$ RHS, assign $0,1,2,3$ to $b$ respectively and pass to next level;



      (Note: $p(9,2),p(8,2)$ etc appears on RHS multiple times, the intention of a dynamic programming is to process these repetition only once.)



      $cdots$



      $p(10,2)=p(10,1)+p(9,1)+cdots+p(0,1)=11$, assign $0,1,2,cdots,10$ to $c$ respectively. And each represent a final scheme, so we have $11$ ways...



      Make a table and do it manually. It's a good educational experience to understand dynamic programming.






      share|cite|improve this answer




























        1














        I am assuming $a,b,c,d in mathbb{Z}$ and non-negative. And even though you mentioned combination, I assume $a=b=c=0,d=10$ is different from $a=b=d=0, c=10$. It very easy to tune it if you intended otherwise.



        This problem can be most elegantly solved with additive number theory method. As I don't know it (yet :), I will approach it with a computer algorithm method called dynamic programming.



        Define $p(n,m)$ the number of ways of partitioning an integer $n$ into sum of $m$ non-negative integers. $p(n,m)$ can be arranged into a $ntimes m$ matrix/grid to minimize counting work.



        So our problem becomes evaluation of $p(10,4)$ with the restriction $a<3, b<4$.



        $p(10,4)=p(10,3)+p(9,3)+p(8,3)$ RHS, assign $0,1,2$ to $a$ respectively and pass to next level



        $p(10,3)=p(10,2)+p(9,2)+p(8,2)+p(7,2)$ RHS, assign $0,1,2,3$ to $b$ respectively and pass to next level；



        $p(9,3)=p(9,2)+p(8,2)+p(7,2)+p(6,2)$ RHS, assign $0,1,2,3$ to $b$ respectively and pass to next level;



        (Note: $p(9,2),p(8,2)$ etc appears on RHS multiple times, the intention of a dynamic programming is to process these repetition only once.)



        $cdots$



        $p(10,2)=p(10,1)+p(9,1)+cdots+p(0,1)=11$, assign $0,1,2,cdots,10$ to $c$ respectively. And each represent a final scheme, so we have $11$ ways...



        Make a table and do it manually. It's a good educational experience to understand dynamic programming.






        share|cite|improve this answer


























          1












          1








          1






          I am assuming $a,b,c,d in mathbb{Z}$ and non-negative. And even though you mentioned combination, I assume $a=b=c=0,d=10$ is different from $a=b=d=0, c=10$. It very easy to tune it if you intended otherwise.



          This problem can be most elegantly solved with additive number theory method. As I don't know it (yet :), I will approach it with a computer algorithm method called dynamic programming.



          Define $p(n,m)$ the number of ways of partitioning an integer $n$ into sum of $m$ non-negative integers. $p(n,m)$ can be arranged into a $ntimes m$ matrix/grid to minimize counting work.



          So our problem becomes evaluation of $p(10,4)$ with the restriction $a<3, b<4$.



          $p(10,4)=p(10,3)+p(9,3)+p(8,3)$ RHS, assign $0,1,2$ to $a$ respectively and pass to next level



          $p(10,3)=p(10,2)+p(9,2)+p(8,2)+p(7,2)$ RHS, assign $0,1,2,3$ to $b$ respectively and pass to next level；



          $p(9,3)=p(9,2)+p(8,2)+p(7,2)+p(6,2)$ RHS, assign $0,1,2,3$ to $b$ respectively and pass to next level;



          (Note: $p(9,2),p(8,2)$ etc appears on RHS multiple times, the intention of a dynamic programming is to process these repetition only once.)



          $cdots$



          $p(10,2)=p(10,1)+p(9,1)+cdots+p(0,1)=11$, assign $0,1,2,cdots,10$ to $c$ respectively. And each represent a final scheme, so we have $11$ ways...



          Make a table and do it manually. It's a good educational experience to understand dynamic programming.






          share|cite|improve this answer














          I am assuming $a,b,c,d in mathbb{Z}$ and non-negative. And even though you mentioned combination, I assume $a=b=c=0,d=10$ is different from $a=b=d=0, c=10$. It very easy to tune it if you intended otherwise.



          This problem can be most elegantly solved with additive number theory method. As I don't know it (yet :), I will approach it with a computer algorithm method called dynamic programming.



          Define $p(n,m)$ the number of ways of partitioning an integer $n$ into sum of $m$ non-negative integers. $p(n,m)$ can be arranged into a $ntimes m$ matrix/grid to minimize counting work.



          So our problem becomes evaluation of $p(10,4)$ with the restriction $a<3, b<4$.



          $p(10,4)=p(10,3)+p(9,3)+p(8,3)$ RHS, assign $0,1,2$ to $a$ respectively and pass to next level



          $p(10,3)=p(10,2)+p(9,2)+p(8,2)+p(7,2)$ RHS, assign $0,1,2,3$ to $b$ respectively and pass to next level；



          $p(9,3)=p(9,2)+p(8,2)+p(7,2)+p(6,2)$ RHS, assign $0,1,2,3$ to $b$ respectively and pass to next level;



          (Note: $p(9,2),p(8,2)$ etc appears on RHS multiple times, the intention of a dynamic programming is to process these repetition only once.)



          $cdots$



          $p(10,2)=p(10,1)+p(9,1)+cdots+p(0,1)=11$, assign $0,1,2,cdots,10$ to $c$ respectively. And each represent a final scheme, so we have $11$ ways...



          Make a table and do it manually. It's a good educational experience to understand dynamic programming.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 24 at 20:04

























          answered Nov 24 at 19:55









          Lance

          54229




          54229























              0














              There are three possible values $(0,1,2)$ for $a$ and four $(0,1,2,3)$ for $b$. Thus there are twelve settings for $a$ and $b$. For each setting, there are $10 - (a+b) + 1$ possible values for $c$. Then $d$ is constrained and fixed.



              So go through the $12$ settings for $a$ and $b$, and for each, add up the number of settings for $c$. And you're done!



              For instance: $a=0, b=0$ leads to $11$ possible values for $c$ $(0, 1, ldots, 10)$ and then $d$ is constrained.



              I get $102$.






              share|cite|improve this answer




























                0














                There are three possible values $(0,1,2)$ for $a$ and four $(0,1,2,3)$ for $b$. Thus there are twelve settings for $a$ and $b$. For each setting, there are $10 - (a+b) + 1$ possible values for $c$. Then $d$ is constrained and fixed.



                So go through the $12$ settings for $a$ and $b$, and for each, add up the number of settings for $c$. And you're done!



                For instance: $a=0, b=0$ leads to $11$ possible values for $c$ $(0, 1, ldots, 10)$ and then $d$ is constrained.



                I get $102$.






                share|cite|improve this answer


























                  0












                  0








                  0






                  There are three possible values $(0,1,2)$ for $a$ and four $(0,1,2,3)$ for $b$. Thus there are twelve settings for $a$ and $b$. For each setting, there are $10 - (a+b) + 1$ possible values for $c$. Then $d$ is constrained and fixed.



                  So go through the $12$ settings for $a$ and $b$, and for each, add up the number of settings for $c$. And you're done!



                  For instance: $a=0, b=0$ leads to $11$ possible values for $c$ $(0, 1, ldots, 10)$ and then $d$ is constrained.



                  I get $102$.






                  share|cite|improve this answer














                  There are three possible values $(0,1,2)$ for $a$ and four $(0,1,2,3)$ for $b$. Thus there are twelve settings for $a$ and $b$. For each setting, there are $10 - (a+b) + 1$ possible values for $c$. Then $d$ is constrained and fixed.



                  So go through the $12$ settings for $a$ and $b$, and for each, add up the number of settings for $c$. And you're done!



                  For instance: $a=0, b=0$ leads to $11$ possible values for $c$ $(0, 1, ldots, 10)$ and then $d$ is constrained.



                  I get $102$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 24 at 20:30

























                  answered Nov 24 at 19:43









                  David G. Stork

                  9,54721232




                  9,54721232






























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