<s>[INST] Please help to find term's coefficient in the following example I trying find the number of all solutions in the following:
$ x_1 + x_2 + x_3 + x_4 + x_5 = 24 $
where: 
2 of variables are natural odd numbers
3 of variables are natural even numbers
none of variables are equal to $0$ or $1$
(all the variables are $>= 2$)
So I've made generating functions according to the limitations as follow:
$ (x^3 + x^5 + x^7 + ... + x^{21} + x^{23})^2 $
$ (x^2 + x^4 + x^6 + ... + x^{22} + x^{24})^3 $
Now I need to find the coefficient of $ x^{24} $
I've tried to solve it but get stuck, here is what I've done:
$ x^6(1 + x^2 + x^4 + ... + x^{18} + x^{20})^2 $
$ x^6(1 + x^2 + x^4 + ... + x^{20} + x^{22})^3 $
$ x^{12}(1 + x^2 + x^4 + ... + x^{18} + x^{20})^2(1 + x^2 + x^4 + ... + x^{20} + x^{22})^3 $
Now we actually looking for coefficient of $ x^{24-12}=x^{12} $ in:
$ (1 + x^2 + x^4 + ... + x^{18} + x^{20})^2(1 + x^2 + x^4 + ... + x^{20} + x^{22})^3 $
$ (\frac{1 - x^{21}}{1 - x})^2(\frac{1 - x^{23}}{1 - x})^3 $
Here I'm stuck. How to proceed in order to find coefficient of $ x^{12} $?
Regards.
PS: Solution was given but just out of curiosity, how to solve it using generating-functions?
 [/INST] Let's take it from the point where we ask for the coefficient of $x^{12}$ in:
$$ (1 + x^2 + x^4 + ... + x^{18} + x^{20})^2(1 + x^2 + x^4 + ... + x^{20} + x^{22})^3 $$
[Note, however, that a factor of $\binom{5}{2}=10$ is already left out of the calculation at this point, which would account for varying positions of the two odd summands $x_i$.  See the comments on Ashtay's Answer for more details.]
Replace $x^2$ with $y$ and ask for the coefficient of $y^6$:
$$ (1 + y + y^2 + \ldots + y^9 + y^{10})^2 (1 + y + y^2 + \ldots + y^{10} + y^{11})^3 $$
Of course we can remove all polynomial in $y$ terms with exponents greater than $6$ since these cannot contribute to the coefficient of $y^6$, then notice the polynomial factors are now equal:
$$ (1 + y + y^2 + \ldots + y^6)^5 $$
We can continue to explicitly work out the coefficient of $y^6$ by repeated squarings (and removing terms beyond $y^6$):
$$ (1 + y + \ldots + y^6)^2 \; \to \; (1 + 2y + 3y^2 + 4y^3 + 5y^4 + 6y^5 + 7y^6) $$
$$ (1 + 2y + 3y^2 + \ldots + 7y^6)^2 \; \to \; (1 + 4y + 10y^2 + 20y^3 + 35y^4 + 56y^5 + 84y^6) $$
Multiplication by $(1+y+y^2+\ldots + y^6)$ of the last result gives the desired term $210y^6$.  It follows that $210$ was the original coefficient of $x^{24}$ you wanted (if my arithmetic is right!).
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