‡ A deck of tri-color cards consists of 32 cards, with 10 cards of each color (red, yellow, blue), 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 the score is calculated as follows: each card with a number $k$ is worth $2^k$ points. If the total score of the drawn cards is 2004, they are called a "good" card group.
Determine the number of "good" card groups.
(Supplied by Tao Pingsheng)
Solution 1: Call the original problem the "Two Kings Problem". If we add another wild card (called the "Middle King"), also numbered 0, and then consider the same problem, it is referred to as the "Three Kings Problem".

First, consider the "Three Kings Problem". Let the large, medium, and small wild cards be called 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, a Yellow Group, and a Blue Group, with the total scores of the Red Group, Yellow Group, and Blue Group denoted as $x$, $y$, and $z$, respectively, such that
$$
x+y+z=n.
$$

Since the binary representation of any non-negative integer is unique, once the values of $x$, $y$, and $z$ are determined, the composition of the Red Group, Yellow Group, and Blue Group is uniquely determined. We know that the number of non-negative integer solutions to the equation $x+y+z=n$ is $\mathrm{C}_{n+1}^{2}$, so
$$
u_{n}=\mathrm{C}_{n+1}^{2}=\frac{(n+1)(n+2)}{2}.
$$

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: (i) If the group contains no wild cards, then the 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 is divided by 2, it becomes a card group in the "Three Kings Problem" with a total score of $k$ and allows for the inclusion of wild cards. It is easy to see that this correspondence is one-to-one, so the number of such card groups is $u_{k}$. (ii) 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
$$
a_{2004}=1003^{2}=1006009.
$$

Solution 2 (Generating Function Method): 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
$$
f(x)=\left(1+x^{2^{0}}\right)^{2}\left(1+x^{2^{1}}\right)^{3} \cdots\left(1+x^{2^{10}}\right)^{3}
$$
(assuming $|x|<1$). Since
$$
\begin{aligned}
f(x) & =\frac{1}{1+x}\left\{\left(1+x^{2^{0}}\right)\left(1+x^{2^{1}}\right)\left(1+x^{2^{2}}\right) \cdots\left(1+x^{2^{10}}\right)\right\}^{3} \\
& =\frac{1}{(1+x)(1-x)^{3}}\left(1-x^{2^{11}}\right)^{3} \\
& =\frac{1}{\left(1-x^{2}\right)(1-x)^{2}}\left(1-x^{2^{11}}\right)^{3},
\end{aligned}
$$

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

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
$$
a_{2004}=1003^{2}=1006009.
$$