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How many solutions possible for the equation$2{x_{1}}+2{x_{2}}+{x_{3}}+{x_{4}}={12}$ all x are non-negative integer.

I see these links but I don't know how to solve this problem.(I know how to solve ${x_1}+{x_2}+{x_3}+{x_4}=12$)

How many solutions possible for the equation $x_1+x_2+x_3+x_4+x_5=55$ if

Enumerating number of solutions to an equation

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    Summing up and case-working is inelegant but I think there is no other way.2012-12-14

2 Answers 2

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Solve for $2a+b=12$. For each one of the $7$ solutions $(a,b)=(k_1,k_2)$, solve for possibilities which count $(k_1+1)\times(k_2+1)$.

Eg. A solution to $2a+b=12$ is $(a,b)=(5,2)$.

So, $x_1+x_2=5$ while $x_3+x_4=2$. Within themselves;

$(x_1,x_2)=(0,5)$,

$(x_1,x_2)=(1,4)$,

...

$(x_1,x_2)=(5,0)$

while

$(x_3,x_4)=(0,2)$,

$(x_3,x_4)=(1,1)$,

$(x_3,x_4)=(2,0)$

adding up to $(5+1)\times(2+1)=18$ solutions for $(a,b)=(5,2)$.

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    we have $(0,12)(1,10)(2,8)(3,6)(4,4)(5,2)(6,0)$ for 2a+b=12 and final answer is 13+22+27+28+25+18+7=140. Thanks.2012-12-15
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Solve $x_{1}+x_{2}+x_{3}+x_{4}=12-x_{1}-x_{2}$ for all natural $x_{1},x_{2}$ s.t $x_{1}+x_{2}\leq6$ s.t and sum the results.

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    @belgi I $w$rite ans$w$er $w$ith your help. can you check ? is it true?2012-12-14