Example 2 A deck of cards consists of 52 cards, with 13 cards of each of the four suits: "Diamonds," "Clubs," "Hearts," and "Spades," numbered $2,3, \cdots, 10, \mathrm{~J}, \mathrm{Q}, \mathrm{K}, \mathrm{A}$, respectively. Two cards of the same suit and adjacent numbers are called "consecutive suited cards," and $\mathrm{A}$ can be considered as 1, making it consecutive with 2. Determine the number of ways to draw 13 cards from this deck such that each number appears and no two cards form a consecutive suited pair.
Solve the problem by first generalizing it as follows:
Let $n \geqslant 3$. From
$$
\begin{array}{l}
A=\left(a_{1}, a_{2}, \cdots, a_{n}\right), B=\left(b_{1}, b_{2}, \cdots, b_{n}\right), \\
C=\left(c_{1}, c_{2}, \cdots, c_{n}\right), D=\left(d_{1}, d_{2}, \cdots, d_{n}\right)
\end{array}
$$

select $n$ terms from these four sequences, satisfying:
(1) Each index $1,2, \cdots, n$ appears;
(2) Any two terms with adjacent indices do not come from the same sequence (indices $n$ and 1 are considered adjacent).
Let the number of selection methods be $x_{n}$, and determine the expression for $x_{n}$.
As shown in Figure 1, divide a circular disk into $n$ sectors, numbered $1,2, \cdots, n$ in sequence, and color the sequences $A, B, C, D$ with four different colors. For any selection scheme, if the term with index $i$ is from a certain colored sequence, then the $i$-th sector is colored with that color.

Thus, $x_{n}$ is the number of ways to color the $n$ sectors of the circular disk with four colors such that no two adjacent sectors have the same color. It is easy to see that,
$$
\begin{array}{l}
x_{1}=4, x_{2}=12, \\
x_{n}+x_{n-1}=4 \times 3^{n-1}(n \geqslant 3) .
\end{array}
$$

Rewrite equation (1) as
$$
\begin{array}{l}
(-1)^{n} x_{n}-(-1)^{n-1} x_{n-1}=-4(-3)^{n-1} \text {. } \\
\text { Then }(-1)^{n-1} x_{n-1}-(-1)^{n-2} x_{n-2}=-4(-3)^{n-2} \text {, } \\
\cdots \cdots \\
(-1)^{3} x_{3}-(-1)^{2} x_{2}=-4(-3)^{2}, \\
(-1)^{2} x_{2}=-4(-3) .
\end{array}
$$

Adding them up, we get
$$
\begin{array}{l}
(-1)^{n} x_{n}=(-3)^{n}+3 \\
\Rightarrow x_{n}=3^{n}+3(-1)^{n}(n \geqslant 2) .
\end{array}
$$

Therefore, $x_{13}=3^{13}-3$, which is the number of card selection methods.

The following problem is similar to Example 2 and can also be solved using the same method.