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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

Community
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Sali Me
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2 Answers2

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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.

Belgi
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  • Can you explain the statement $12-x_1-x_2$ and $ x_1+x_2 \le 6 $ a bit more? and continue more? – Sali Me Dec 14 '12 at 12:21
  • If $x_1+x_2\geq 6$ then $2x_1+2x_2>12$ and since $x_3,x_4$ are natural there are no solutions to the equations. Fix any pair $x_1,x_2$ s.t their sum is $\leq 6$ (hence there are solutions) and find how many $x_3,x_4$ there are that solve the equation. Note that for such fixed $x_1,x_2$ you want to count the number of solutions for $x_3+x_4=12-2x_1-2x_2$ which you already know to solve – Belgi Dec 14 '12 at 12:27
  • It seems you've somehow added an $x_5$ that isn't in the question. – joriki Dec 14 '12 at 12:29
  • @joriki - thanks! I edited accordingly – Belgi Dec 14 '12 at 12:31
  • @belgi I write answer with your help. can you check ? is it true? – Sali Me Dec 14 '12 at 12:59
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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 $(a,b)=(5,2)$.

ashley
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