<s>[INST] What is the number of integer solutions of $2{x_{1}}+2{x_{2}}+{x_{3}}+{x_{4}}={12}$? What is the number of non negative integer solutions of $2{x_{1}}+2{x_{2}}+{x_{3}}+{x_{4}}={12}$ ? 

I tried it as :
${x_{3}}+{x_{4}}={12} - 2{x_{1}}-2{x_{2}}$
Now, finding the solutions of ${x_{1}}+{x_{2}}$


*

*${x_{1}}+{x_{2}} = 0 => 1$ Solution

*${x_{1}}+{x_{2}} = 1 => 2$ Solutions

*${x_{1}}+{x_{2}} = 2 => 3$ Solutions

*${x_{1}}+{x_{2}} = 3 => 4$ Solutions

*${x_{1}}+{x_{2}} = 4 => 5$ Solutions

*${x_{1}}+{x_{2}} = 5 => 6$ Solutions

*${x_{1}}+{x_{2}} = 6 => 7$ Solutions


And, now respectively finding for ${x_{3}}+{x_{4}}$


*

*${x_{3}}+{x_{4}} = 12 => 13$ Solutions

*${x_{3}}+{x_{4}} = 10 => 11$ Solutions

*${x_{3}}+{x_{4}} = 8 => 9$ Solutions

*${x_{3}}+{x_{4}} = 6 => 7$ Solutions

*${x_{3}}+{x_{4}} = 4 => 5$ Solutions

*${x_{3}}+{x_{4}} = 2 => 3$ Solutions

*${x_{3}}+{x_{4}} = 0 => 1$ Solution


Then, Multiplying respective numbers
$1.13 +  2.11 + 3.9 + 4.7 + 5.5 + 6.3 + 7.1 = 140$ Solutions

I don't have answer for it. Am i right here ?
 [/INST] Here is another solution, using generating formulas:
Let the power of $x$ represent the value of $x_i$.
Then we have $$(1+x^2+x^4+\cdots)\times (1+x^2+x^4+\cdots)\times(1+x+x^2+\cdots)\times (1+x+x^2+\cdots)$$
$$=\left(\frac{1}{1-x^2}\right)^2\cdot\left(\frac{1}{1-x}\right)^2$$
$$=\frac{1}{(1-x^2)^2(1-x)^2}$$
$$=\frac{1}{1-2x-x^2+4x^3-x^4-2x^5+x^6}.$$
Now we want to find the coefficient of $x^{12}$. By (arduous or computer aided) long division, we get the same answer, $140$.
By the way, one easy way to do long division here is to type the expression into Wolfram Alpha with the phrase "taylor series for." Then it will give you the option to calculate more terms.
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