Number of solutions to $x_1+x_2+\cdots+x_5=41$ 
Use a generating function to count the number of integer solutions
   to $x_1+x_2+\cdots+x_5=41$ that satisfy $0\le x_i\le20$ for all $i$,
   $x_i$ is even when $i$ is even, and $x_i$ is odd when $i$ is odd.

Ignoring the conditions at the end, I believe that $g(x)=(1+x+x^2+\cdots+x^{20})^5$, and the answer to the problem is the coefficient of $x^{41}$ in $g(x)$. But, as I said, I have ignored the terms after all; I haven't the faintest idea of how to go about incorporating them. I'm (obviously) missing something, and I don't seem to be getting anywhere, no matter how many examples I go through. Any help is greatly appreciated!!
 A: You're a good deal of the way there. What you have so far is five copies of $1+x+\cdots+x^{20}$. They all stop at $20$ because no $x_i$ can be larger than $20$.
The limitation that $x_1$ can only be odd is similarily simple. It just means that the factor of the generating function that corresponds to $x_1$ only has odd-degree terms. So instead of $1+x+x^2+\cdots + x^{20}$, you get $x + x^3 + \cdots + x^{19}$. Do a similar thing for the other four variables, and you have your generating function.
A: In the general scenario,
$$x_1 + x_2 + ... + x_n = r$$
where $x_k \geq 0$ for all $k$, we have a generating function $g$ that is a product of $n$ polynomials $P$:
$$g(x) = P_1(x)P_2(x)...P_n(x)$$
Each of these $P_k(x)$ is defined by the restrictions on the $x_k$. Each $P_k$ will be a sum of powers of $x$, where each exponent of a $x$ term is a valid value $x_k$ may take on in the above equation. Some examples are prudent:


Example $\#1$: We want to find the number of nonnegative integer solutions to
  $$x_1 + x_2 + x_3 = 10$$
  where $x_i \geq 0$ for all $i$

This is the scenario you seem to be fine with: effectively no restrictions. Though it must be noted that there is an implicit maximum of $10$ here for each $x_k$: as you deal with more of these, you will learn whether to use $0 \le x_k \le 10$ or simply $0 \le x_k$ (and thus use an infinite sum). I won't bog us down on the details here regarding which to use, and simply assume finite sums.
Here, since $x_k$ can be $0,1,...,10$, then, each of the $P_k$ polynomials are
$$P_k(x) = 1 + x + x^2 + x^3 + ... + x^{10}$$
(Notice: $x^0 = 1$.) Then the generating function is the product of $P_1,P_2,P_3$ - which are all the same and thus give you the cube of the above.


Example $\#2$: We want to find the number of nonnegative integer solutions to
  $$x_1 + x_2 + x_3 = 10$$
  where $x_i \geq 0$ for all $i$, and $x_i$ must be even.

In this scenario, the allowed values for $x_i$ have changed. Now we cannot have odd values! Thus, $x_k$ must be $0,2,4,6,8,$ or $10$ for all $k$. Since these are the allowed values, we see that
$$P_1(x) = P_2(x) = P_3(x) = 1 + x^2 + x^4 + x^6 + x^8 + x^{10}$$
Again, bear in mind how the exponents (when $1 = x^0$) correspond to the allowed values for $x_k$!
Again in this case, $P_1 = P_2 = P_3$, so $g$ is just the cube of any one of them.


Example $\#3$: We want to find the number of nonnegative integer solutions to
  $$x_1 + x_2 + x_3 = 10$$
  where $x_i \geq 0$ for all $i$. Further, $x_1$ must be even, $x_2$ must be prime, and $x_3$ must be a perfect square.

Now have different conditions for all three numbers! But this is still doable! We first individually see what values are allowed per variable. On investigation...


*

*$x_1$ can be $0,2,4,6,8,10$

*$x_2$ can be $2,3,5,7$

*$x_3$ can be $0,1,4,9$
With these in mind, we construct the polynomials one by one. Again, the exponents are the permissibile powers of the corresponding summand, with $x_i^0 = 1$. Then we see


*

*$P_1(x) = 1 + x^2 + x^4 + x^6 + x^8 + x^{10}$

*$P_2(x) = x^2 + x^3 + x^7 + x^9$

*$P_3(x) = 1 + x + x^4 + x^9$
Our generating function, $g(x)$, this time is not a cube, but simply the product
$$g(x) = P_1(x)P_2(x)P_3(x)$$
If one wanted to write it explicitly they could with no trouble, just substitute in the expressions in the bullets.

With these in mind, we consider now your scenario:
$$x_1 + x_2 + x_3 + x_4 + x_5 = 41$$
where, for $k = 1,2,...,5$,


*

*$x_k \ge 0$

*$x_k \le 20$

*$x_k$ is even for even $k$

*$x_k$ is odd for odd $k$
We translate this into the sets of permissible values. Keep in mind that we cannot go any higher than $41$ necessarily for any value. If you want to, I think this is the kind of equation that would allow for an infinite sum instead of finite - and if you're familiar with that, the work is similar enough that you should be able to analogize it. 
In any event, the set of permissible values:


*

*$x_2,x_4$ can be $0,2,4,6,...,20$

*$x_1,x_3,x_5$ can be $1,3,5,7,...,19$
Then the corresponding polynomials are? We again construct them by basing the exponents on the allowed values:
$$\begin{align}
P_1(x) = P_3(x) = P_5(x) &= x + x^3 + x^5 + x^7 + ... + x^{19}\\
P_2(x) = P_4(x) &= 1 + x^2 + x^4 + x^6 + ... + x^{20}
\end{align}$$
and then $g(x) = P_1(x)P_2(x)P_3(x)P_4(x)P_5(x)$.
