1. There are 10 balls each of red, white, and black. If 12 balls are drawn from them, but it is required that all three colors of balls are included, how many different ways are there to draw the balls?
1. Solution 1 Since balls of the same color are indistinguishable, it belongs to a 12-element repetition combination of 3 elements. Also, because each draw requires balls of all three colors, we can first take one red, one white, and one black ball, and then take 9 balls without restriction from the red, white, and black balls, with each color of ball being able to take $0,1,2, \cdots, 9$ balls. Therefore, if we set the total number of ways to draw $r$ balls as $a_{r}$, the generating function for $\left\{a_{r}\right\}$ is $f(x)=\left(1+x+x^{2}+\cdots+x^{9}\right)^{3}$.
Since $f(x)=\left(\frac{1-x^{10}}{1-x}\right)^{3}=\left(1-x^{10}\right)^{3} \cdot(1-x)^{-3}=\left(1-x^{10}\right)^{3} \sum_{r=0}^{\infty} C_{r+2}^{r} x^{r}$.
Clearly, the coefficient of $x^{9}$ can only be obtained by setting $r=9$ within the $\sum$. Hence, $a_{9}=C_{11}^{9}=55$ is the answer.
Solution 2 According to the problem, at least one ball of each color must be drawn, meaning no color can be drawn 0 times, and the maximum number of balls of each color that can be drawn is 10. Therefore, if we set the total number of ways to draw $r$ balls as $a_{r}$, the generating function for $\left.\mid a_{r}\right\}$ is $f(x)=\left(x+x^{2}+\cdots+x^{10}\right)^{3}$.
$$
\text { By } \begin{aligned}
f(x) & =x^{3}\left(1+x+x^{2}+\cdots+x^{9}\right)^{3}=x^{3} \cdot\left(\frac{1-x^{10}}{1-x}\right)^{3}=x^{3} \cdot\left(1-x^{10}\right)^{3} \cdot(1-x)^{-3} \\
& =x^{3}\left(1-3 x^{10}+3 x^{20}-x^{30}\right) \cdot \sum_{r=0}^{\infty} C_{r+2}^{r} x^{r} .
\end{aligned}
$$

Clearly, the coefficient of $x^{12}$ can only be obtained by setting $r=9$ within the $\sum$. Hence, $a_{12}=C_{11}^{9}=55$ is the answer.