Example 6 The number of non-negative integer solutions to the equation $2 x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}+x_{7}+x_{8}+x_{9}+x_{10}=3$ is?
(1985 National High School League Question)
Solution 1 The generating function is $\frac{1}{\left(1-x^{2}\right)(1-x)^{9}}=\frac{1}{1+x} \cdot \frac{1}{(1-x)^{10}}=$
$$
\frac{A}{1+x}+\frac{B}{1-x}+\frac{C}{(1-x)^{2}}+\frac{D}{(1-x)^{3}}+\cdots+\frac{N}{(1-x)^{10}},
$$

Using the method of undetermined coefficients, we can find $A, B, C, D, \cdots, N$, so
$$
\begin{array}{l} 
\frac{1}{\left(1-x^{2}\right)(1-x)^{9}}= \frac{1}{1024(1+x)}+\frac{1}{1024(1-x)}+\frac{1}{512(1-x)^{2}}+\frac{1}{256(1-x)^{3}}+\cdots+ \\
\frac{1}{2(1-x)^{10}} \\
= \sum_{r=0}^{\infty}\left[\frac{1}{1024}(-1)^{r}+\frac{1}{1024} C_{r}^{r}+\frac{1}{256} C_{r+2}^{r}+\frac{1}{128} C_{r+3}^{r}+\frac{1}{64} C_{r+4}^{r}+\right. \\
\left.\frac{1}{32} C_{r+5}^{r}+\cdots+\frac{1}{2} C_{r+9}^{r}\right] x^{r} .
\end{array}
$$

Therefore, when $r=3$, the coefficient of $x^{3}$ is $\frac{89088}{512}=174$.
Solution 2 Since $x_{1}$ can only take 0 or 1, when $x_{1}=0$, the original indeterminate equation becomes $x_{2}+x_{3}+\cdots+$ $x_{10}=3$, the generating function for its non-negative integer solutions is $\frac{1}{(1-x)^{9}}=\sum_{r=0}^{\infty} C_{9+r-1}^{r} x^{r}$, at this time $r=3$, so the number of non-negative integer solutions is $C_{9+3-1}^{3}=165$; when $x_{1}=1$, the original indeterminate equation becomes $x_{2}+x_{3}+\cdots+x_{10}=1$, the generating function for its non-negative integer solutions is $\frac{1}{(1-x)^{9}}=\sum_{r=0}^{\infty} C_{9+r-1}^{r} x^{r}$, at this time $r=1$, so the number of non-negative integer solutions is $C_{9+1-1}^{1}=9$. Therefore, the number of non-negative integer solutions to the original indeterminate equation is $165+9=174$.