counting ways for sum with maximums on variables
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
edited Nov 25 at 11:29
N. F. Taussig
43.5k93355
asked Nov 24 at 19:29
WEPIOD
31
add a comment |
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
edited Nov 25 at 11:29
N. F. Taussig
43.5k93355
asked Nov 24 at 19:29
WEPIOD
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
add a comment |
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
edited Nov 25 at 11:29
N. F. Taussig
43.5k93355
asked Nov 24 at 19:29
WEPIOD
31
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
edited Nov 25 at 11:29
N. F. Taussig
43.5k93355
asked Nov 24 at 19:29
WEPIOD
31
edited Nov 25 at 11:29
N. F. Taussig
43.5k93355
asked Nov 24 at 19:29
WEPIOD
31
edited Nov 25 at 11:29
N. F. Taussig
43.5k93355
edited Nov 25 at 11:29
N. F. Taussig
43.5k93355
edited Nov 25 at 11:29
N. F. Taussig
43.5k93355
43.5k93355
asked Nov 24 at 19:29
WEPIOD
31
asked Nov 24 at 19:29
WEPIOD
31
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
add a comment |
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
add a comment |
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.
answered Nov 24 at 19:55
Lance
54229
add a comment |
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 $102$.
answered Nov 24 at 19:43
David G. Stork
9,54721232
add a comment |
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2 Answers
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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.
answered Nov 24 at 19:55
Lance
54229
add a comment |
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.
answered Nov 24 at 19:55
Lance
54229
add a comment |
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.
answered Nov 24 at 19:55
Lance
54229
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.
answered Nov 24 at 19:55
Lance
54229
edited Nov 24 at 20:04
answered Nov 24 at 19:55
Lance
54229
answered Nov 24 at 19:55
Lance
54229
answered Nov 24 at 19:55
Lance
54229
54229
add a comment |
add a comment |
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 $102$.
answered Nov 24 at 19:43
David G. Stork
9,54721232
add a comment |
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 $102$.
answered Nov 24 at 19:43
David G. Stork
9,54721232
add a comment |
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 $102$.
answered Nov 24 at 19:43
David G. Stork
9,54721232
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 $102$.
answered Nov 24 at 19:43
David G. Stork
9,54721232
edited Nov 24 at 20:30
answered Nov 24 at 19:43
David G. Stork
9,54721232
answered Nov 24 at 19:43
David G. Stork
9,54721232
answered Nov 24 at 19:43
David G. Stork
9,54721232
9,54721232
add a comment |
add a comment |
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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.
– 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