number of solutions to $x_1 + x_2 + x_3 + x_4 + x_5 = 31$ via generating function? I will be very happy to understand how to solve this problem with generating function:
How many solutions are there to the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 = 31$$
where $x_i$ is a nonnegative integer, 
and $x_2$ is even and $x_3$ is odd?
thanks!
 A: Non-negative $x_1$ corresponds to $\frac{z}{1 - z}$, $x_2$ even gives rise to $\frac{1}{1 - z^2}$, odd $x_3$ means $\frac{z}{1 - z^2}$, non-restriced $x_4$, $x_5$ are $\frac{1}{1 - z}$ each:
\begin{align}
[z^{31}] &\frac{z}{1 - z} 
           \cdot \frac{1}{1 - z^2}
           \cdot \frac{z}{1 - z^2}
           \cdot \frac{1}{(1 - z)^2} \\
  &= [z^{31}] \frac{z^2}{(1 - z)^3 (1 - z^2)^2} \\
  &= [z^{29}] \left(
                \frac{5}{64 (1 + z)}
                  + \frac{1}{32 (1 + z)^2}
                  + \frac{5}{64 (1 - z)}
                  + \frac{1}{8 (1 - z)^2}
                  + \frac{3}{16 (1 - z)^3}
                  + \frac{1}{4 (1 - z)^4}
                  + \frac{1}{4 (1 - z)^5}
              \right)
\end{align}
Use the fact that $\binom{-n}{k} = (-1)^k \binom{k + n - 1}{n - 1}$:
\begin{align}
[z^{31}] &\frac{z^2}{1 - z} 
           \cdot \frac{1}{1 - z^2}
           \cdot \frac{z}{1 - z^2}
           \cdot \frac{1}{(1 - z)^2} \\
          &= \frac{5}{64} \binom{-1}{29}
               + \frac{1}{32} \binom{-2}{29}
               + \frac{5}{64} \binom{-1}{29} (-1)^{29}
               + \frac{1}{8} \binom{-2}{29} (-1)^{29}
               + \frac{3}{16} \binom{-3}{29} (-1)^{29}
               + \frac{1}{4} \binom{-4}{29} (-1)^{29}
               + \frac{1}{4} \binom{-5}{29} (-1)^{29} \\
          &= 11560
\end{align}
A: The generating function is $(1/(1-x))^3 \times 1/(1-x^2) \times x/(1-x^2)$. (Why?) The coefficient of $x^{31}$ in the series expansion gives you your answer. 
