Example 25 (2004 National Women's Mathematical Olympiad Problem) A deck of tri-color cards, totaling 32 cards, includes 10 cards of each of the red, yellow, and blue colors, numbered $1, 2, \cdots, 10$; there are also a big joker and a small joker, each numbered 0. From this deck, several cards are drawn, and then the score is calculated according to the following rule: each card numbered $k$ is counted as $2^{k}$ points; if their total score is 2004, then these cards are called a "good card group." Try to find the number of "good card groups."
Solution 1 Change the original problem's statement.
Call the original problem the "Two Kings Problem". If an additional wild card (called the "Middle King") is added, also numbered 0, and the same problem is considered, it is called the "Three Kings Problem".

First, consider the "Three Kings Problem". The large, medium, and small wild cards are referred to as the Red King, Yellow King, and Blue King, respectively. Thus, each color has 11 cards, numbered $0, 1, 2, \cdots, 10$. Let the number of card groups with a total score of $n$ be denoted as $u_{n}$. Each card group can be composed of a Red Group, Yellow Group, and Blue Group. Let the total scores of the Red Group, Yellow Group, and Blue Group be denoted as $x, y, z$, respectively, then
$$
x+y+z=n \text {. }
$$

Since any non-negative integer has a unique binary representation, once the values of $x, y, z$ are determined, the composition of the Red Group, Yellow Group, and Blue Group is uniquely determined. The number of non-negative integer solutions to the equation $x+y+z=n$ is equal to $C_{n+2}^{2}$, so,
$$
u_{n}=C_{n+2}^{2}=\frac{(n+1)(n+2)}{2} \text {. }
$$

Now consider the original "Two Kings Problem". For $n \in\{1,2, \cdots, 2004\}$, let $a_{n}$ denote the number of card groups with a total score of $n$.
When $n=2 k \leqslant 2004$, for any card group with a total score of $2 k$, we have:
(1) If the group contains no wild cards, then this card group is a card group in the "Three Kings Problem" with a total score of $2 k$ and no wild cards. If the score of each card in the group is divided by 2, it becomes a card group in the "Three Kings Problem" with a total score of $k$ and allows for wild cards. It is easy to see that this correspondence is one-to-one, so the number of such card groups is $u_{k}$.
(2) If the group contains wild cards, then the group must contain 2 wild cards (both the large and small wild cards are in the group). Removing the wild cards reduces it to a card group with a total score of $2 k-2$ and no wild cards. Thus, the number of such card groups is $u_{k-1}$. Therefore,
$$
\begin{aligned}
a_{2 k} & =u_{k}+u_{k-1}=\frac{(k+1)(k+2)}{2}+\frac{k(k+1)}{2} \\
& =(k+1)^{2}, k=1,2, \cdots, 1002 .
\end{aligned}
$$

In particular, the number of "good card groups" is
$$
a_{2004}=1003^{2}=1006009 \text {. }
$$

Solution 2 Change the original problem's statement.
For $n \in\{1,2, \cdots, 2004\}$, let $a_{n}$ denote the number of card groups with a total score of $n$, then $a_{n}$ is equal to the coefficient of $x^{n}$ in the expansion of the function
$f(x)=\left(1+x^{2^{0}}\right)^{2}+\left(1+x^{2^{1}}\right)^{3}\left(1+x^{2^{2}}\right)^{3} \cdots\left(1+x^{2^{10}}\right)^{3}$ (with the convention $|x|<1$). Since
$$
\begin{aligned}
f(x) & =\frac{1}{x+1}\left[\left(1+x^{2^{0}}\right)\left(1+x^{2^{1}}\right)\left(1+x^{2^{2}}\right) \cdots\left(1+x^{2^{10}}\right)\right]^{3} \\
& =\frac{1}{(1+x)(1-x)^{3}}\left(1-x^{2^{11}}\right)^{3} \\
& \left.=\frac{1}{\left(1-x^{2}\right)(1-x)^{2}} 1-x^{2^{11}}\right)^{3},
\end{aligned}
$$

and $n \leqslant 2004<2^{11}$, so $a_{n}$ is equal to the coefficient of $x^{n}$ in the expansion of $\frac{1}{\left(1-x^{2}\right)(1-x)^{2}}$. Also, since
$$
\begin{aligned}
\frac{1}{\left(1-x^{2}\right)(1-x)^{2}} & =\frac{1}{1-x^{2}} \cdot \frac{1}{(1-x)^{2}} \\
& =\left(1+x^{2}+x^{4}+\cdots+x^{2 k}+\cdots\right) \cdot\left(1+2 x+3 x^{2}+\cdots+(2 k+1) x^{2 k}+\cdots\right),
\end{aligned}
$$

it follows that the coefficient of $x^{2 k}$ is
$$
a_{2 k}=1+3+5+\cdots+(2 k+1)=(k+1)^{2}, k=1,2 \cdots
$$

Thus, the number of "good card groups" is
$$
a_{2004}=1003^{2}=1006009
$$