How many integer solutions with negative numbers?
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If a question asked: How many integer solutions of $x_1+x_2+x_3+x_4=30$ with $-9 leq x_i leq 21$? How would this be solved?
I understand how to solve if the inequality was $0 leq x_i leq 21$?, but how to solve between a negative and positive inequality?
$N = binom{30 + 4-1}{30}$
$N(A_i) = binom{(30 - ?) +4-1}{30 - ?}$
...
combinatorics
asked Dec 10 '18 at 20:41
Math NewbieMath Newbie
428
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add a comment |
$begingroup$
If a question asked: How many integer solutions of $x_1+x_2+x_3+x_4=30$ with $-9 leq x_i leq 21$? How would this be solved?
I understand how to solve if the inequality was $0 leq x_i leq 21$?, but how to solve between a negative and positive inequality?
$N = binom{30 + 4-1}{30}$
$N(A_i) = binom{(30 - ?) +4-1}{30 - ?}$
...
combinatorics
asked Dec 10 '18 at 20:41
Math NewbieMath Newbie
428
$endgroup$
$begingroup$
It is the coefficient of $x^{30}$ in $$frac{x^{-36}(1-x^{30})^4}{(1-x)^4}$$
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– Matt Samuel
Dec 10 '18 at 20:44
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@MattSamuel Could you explain this? We have to use sets (finished generating functions chapter).
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– Math Newbie
Dec 10 '18 at 20:46
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The formula for $N$ you give is the unrestricted value. That is, it would be the answer if the question was how many solutions are there to $x_1+x_2+x_3+x_4=30$ with $x_i≥0$. More work is needed to handle a cap, as in $0≤x_i≤21$.
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– lulu
Dec 10 '18 at 20:58
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"set theory" has nothing to do with your question.
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– Jean Marie
Dec 10 '18 at 21:47
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@JeanMarie Ah, sorry, I assumed so because that's our current chapter.
$endgroup$
– Math Newbie
Dec 10 '18 at 22:34
add a comment |
$begingroup$
If a question asked: How many integer solutions of $x_1+x_2+x_3+x_4=30$ with $-9 leq x_i leq 21$? How would this be solved?
I understand how to solve if the inequality was $0 leq x_i leq 21$?, but how to solve between a negative and positive inequality?
$N = binom{30 + 4-1}{30}$
$N(A_i) = binom{(30 - ?) +4-1}{30 - ?}$
...
combinatorics
asked Dec 10 '18 at 20:41
Math NewbieMath Newbie
428
$endgroup$
If a question asked: How many integer solutions of $x_1+x_2+x_3+x_4=30$ with $-9 leq x_i leq 21$? How would this be solved?
I understand how to solve if the inequality was $0 leq x_i leq 21$?, but how to solve between a negative and positive inequality?
$N = binom{30 + 4-1}{30}$
$N(A_i) = binom{(30 - ?) +4-1}{30 - ?}$
...
combinatorics
combinatorics
asked Dec 10 '18 at 20:41
Math NewbieMath Newbie
428
asked Dec 10 '18 at 20:41
Math NewbieMath Newbie
428
edited Dec 10 '18 at 22:34
Math Newbie
asked Dec 10 '18 at 20:41
Math NewbieMath Newbie
428
asked Dec 10 '18 at 20:41
Math NewbieMath Newbie
428
asked Dec 10 '18 at 20:41
Math NewbieMath Newbie
428
428
$begingroup$
It is the coefficient of $x^{30}$ in $$frac{x^{-36}(1-x^{30})^4}{(1-x)^4}$$
$endgroup$
– Matt Samuel
Dec 10 '18 at 20:44
$begingroup$
@MattSamuel Could you explain this? We have to use sets (finished generating functions chapter).
$endgroup$
– Math Newbie
Dec 10 '18 at 20:46
$begingroup$
The formula for $N$ you give is the unrestricted value. That is, it would be the answer if the question was how many solutions are there to $x_1+x_2+x_3+x_4=30$ with $x_i≥0$. More work is needed to handle a cap, as in $0≤x_i≤21$.
$endgroup$
– lulu
Dec 10 '18 at 20:58
$begingroup$
"set theory" has nothing to do with your question.
$endgroup$
– Jean Marie
Dec 10 '18 at 21:47
$begingroup$
@JeanMarie Ah, sorry, I assumed so because that's our current chapter.
$endgroup$
– Math Newbie
Dec 10 '18 at 22:34
add a comment |
$begingroup$
It is the coefficient of $x^{30}$ in $$frac{x^{-36}(1-x^{30})^4}{(1-x)^4}$$
$endgroup$
– Matt Samuel
Dec 10 '18 at 20:44
$begingroup$
@MattSamuel Could you explain this? We have to use sets (finished generating functions chapter).
$endgroup$
– Math Newbie
Dec 10 '18 at 20:46
$begingroup$
The formula for $N$ you give is the unrestricted value. That is, it would be the answer if the question was how many solutions are there to $x_1+x_2+x_3+x_4=30$ with $x_i≥0$. More work is needed to handle a cap, as in $0≤x_i≤21$.
$endgroup$
– lulu
Dec 10 '18 at 20:58
$begingroup$
"set theory" has nothing to do with your question.
$endgroup$
– Jean Marie
Dec 10 '18 at 21:47
$begingroup$
@JeanMarie Ah, sorry, I assumed so because that's our current chapter.
$endgroup$
– Math Newbie
Dec 10 '18 at 22:34
$begingroup$
It is the coefficient of $x^{30}$ in $$frac{x^{-36}(1-x^{30})^4}{(1-x)^4}$$
$endgroup$
– Matt Samuel
Dec 10 '18 at 20:44
$begingroup$
It is the coefficient of $x^{30}$ in $$frac{x^{-36}(1-x^{30})^4}{(1-x)^4}$$
$endgroup$
– Matt Samuel
Dec 10 '18 at 20:44
$begingroup$
@MattSamuel Could you explain this? We have to use sets (finished generating functions chapter).
$endgroup$
– Math Newbie
Dec 10 '18 at 20:46
$begingroup$
@MattSamuel Could you explain this? We have to use sets (finished generating functions chapter).
$endgroup$
– Math Newbie
Dec 10 '18 at 20:46
$begingroup$
The formula for $N$ you give is the unrestricted value. That is, it would be the answer if the question was how many solutions are there to $x_1+x_2+x_3+x_4=30$ with $x_i≥0$. More work is needed to handle a cap, as in $0≤x_i≤21$.
$endgroup$
– lulu
Dec 10 '18 at 20:58
$begingroup$
The formula for $N$ you give is the unrestricted value. That is, it would be the answer if the question was how many solutions are there to $x_1+x_2+x_3+x_4=30$ with $x_i≥0$. More work is needed to handle a cap, as in $0≤x_i≤21$.
$endgroup$
– lulu
Dec 10 '18 at 20:58
$begingroup$
"set theory" has nothing to do with your question.
$endgroup$
– Jean Marie
Dec 10 '18 at 21:47
$begingroup$
"set theory" has nothing to do with your question.
$endgroup$
– Jean Marie
Dec 10 '18 at 21:47
$begingroup$
@JeanMarie Ah, sorry, I assumed so because that's our current chapter.
$endgroup$
– Math Newbie
Dec 10 '18 at 22:34
$begingroup$
@JeanMarie Ah, sorry, I assumed so because that's our current chapter.
$endgroup$
– Math Newbie
Dec 10 '18 at 22:34
add a comment |
1 Answer
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$begingroup$
Let $y_i=x_i+9$. Then
$$
y_1+y_2+y_3+y_4=66\
0leq y_ileq 30
$$
Now find the number of solutions the way you say you know.
answered Dec 10 '18 at 20:48
ArthurArthur
114k7115197
$endgroup$
add a comment |
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$begingroup$
Let $y_i=x_i+9$. Then
$$
y_1+y_2+y_3+y_4=66\
0leq y_ileq 30
$$
Now find the number of solutions the way you say you know.
answered Dec 10 '18 at 20:48
ArthurArthur
114k7115197
$endgroup$
add a comment |
$begingroup$
Let $y_i=x_i+9$. Then
$$
y_1+y_2+y_3+y_4=66\
0leq y_ileq 30
$$
Now find the number of solutions the way you say you know.
answered Dec 10 '18 at 20:48
ArthurArthur
114k7115197
$endgroup$
add a comment |
$begingroup$
Let $y_i=x_i+9$. Then
$$
y_1+y_2+y_3+y_4=66\
0leq y_ileq 30
$$
Now find the number of solutions the way you say you know.
answered Dec 10 '18 at 20:48
ArthurArthur
114k7115197
$endgroup$
Let $y_i=x_i+9$. Then
$$
y_1+y_2+y_3+y_4=66\
0leq y_ileq 30
$$
Now find the number of solutions the way you say you know.
answered Dec 10 '18 at 20:48
ArthurArthur
114k7115197
answered Dec 10 '18 at 20:48
ArthurArthur
114k7115197
answered Dec 10 '18 at 20:48
ArthurArthur
114k7115197
answered Dec 10 '18 at 20:48
ArthurArthur
114k7115197
114k7115197
add a comment |
add a comment |
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$begingroup$
It is the coefficient of $x^{30}$ in $$frac{x^{-36}(1-x^{30})^4}{(1-x)^4}$$
$endgroup$
– Matt Samuel
Dec 10 '18 at 20:44
$begingroup$
@MattSamuel Could you explain this? We have to use sets (finished generating functions chapter).
$endgroup$
– Math Newbie
Dec 10 '18 at 20:46
$begingroup$
The formula for $N$ you give is the unrestricted value. That is, it would be the answer if the question was how many solutions are there to $x_1+x_2+x_3+x_4=30$ with $x_i≥0$. More work is needed to handle a cap, as in $0≤x_i≤21$.
$endgroup$
– lulu
Dec 10 '18 at 20:58
$begingroup$
"set theory" has nothing to do with your question.
$endgroup$
– Jean Marie
Dec 10 '18 at 21:47
$begingroup$
@JeanMarie Ah, sorry, I assumed so because that's our current chapter.
$endgroup$
– Math Newbie
Dec 10 '18 at 22:34