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Following my post here: Newton Binomial to find $(x_1+x_2+x_3)(x_4+x_5+x_6+x_7)=77$

I would like to know the number of integer solutions to |x|+|y|+|z| = 30. Without the absolute value I get 32 choose 30 (496). With the absolutes, I get 3602 (by computer simulation) and I don't know why...

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    $\begingroup$ Title: "Neuton" should be "Newton". $\endgroup$ Mar 23 at 15:13
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The formula for the number of positive solutions to the equation $$ x_1+x_2+\dots+x_k=n $$ reads $$ \binom {n-1}{k-1}. $$ Therefore the number your are looking for is: $$ 2^3\binom{29}2+\binom312^2\binom{29}1+\binom322^1\binom{29}0=3602, $$ where the terms count contributions with the number of zeros in the set $\{a,b,c\}$ being $0,1$ and $2$, respectively.

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Hint Consider the equation $$ a+b+c=30 $$ with $0 \leq a,b,c$.

For each solution with $a,b,c \geq 1$ you get $8$ solutions $x= \pm a, y= \pm b, z=\pm c$ to your equation.

For each solution with exactly one of $a,b,c$ equals zero you get $4$ solutions to your equation.

For each solution with exactly one of $a,b,c$ equals zero you get $2$ solutions to your equation.

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  • $\begingroup$ Thanks allot for your answer! Unfortunately I still can't figure out how to incorporate your hint :-( Can you elaborate? Thanks $\endgroup$
    – Avi Tal
    Mar 23 at 15:38
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    $\begingroup$ avi, after this you can use generating functions or stars and bars $\endgroup$ Mar 23 at 16:32
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    $\begingroup$ @AviTal If $a,b,c \geq 1$ then set $a'=a-1,b'=b-1,c'=c-1$ and use stars and bars. If $a=0$ and $b,c >1$ you need to solve $b+c=30$ with $b \geq 1, c \geq 1$. AGain, set $b'=b-1, c'=c-2$ and solve via stars and bars.. $\endgroup$
    – N. S.
    Mar 23 at 20:12

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