Example 1 A deck of cards consists of 52 cards, with 13 cards of each of the four suits: "Diamonds," "Clubs," "Hearts," and "Spades." The cards are numbered $2,3, \cdots, 10, \mathrm{~J}, \mathrm{Q}, \mathrm{K}, \mathrm{A}$. Two cards of the same suit and consecutive numbers are called "straight flush" cards, and $\mathrm{A}$ can be considered as 1, making it a straight flush with 2. Determine the number of ways to draw 13 cards from this deck such that each number appears and no straight flush cards are included.
(3rd China Southeast Mathematical Olympiad)
Solution: First, generalize the problem as follows:
$$
\begin{array}{l}
\text { Let } n \geqslant 3 \text {, from } \\
A=\left(a_{1}, a_{2}, \cdots, a_{n}\right) 、 B=\left(b_{1}, b_{2}, \cdots, b_{n}\right) \text { 、 } \\
C=\left(c_{1}, c_{2}, \cdots, c_{n}\right) 、 D=\left(d_{1}, d_{2}, \cdots, d_{n}\right) \\
\end{array}
$$

Select $n$ terms from these four sequences, satisfying:
(1) Each index $1,2, \cdots, n$ appears;
(2) Any two adjacent terms do not come from the same sequence (index $n$ and 1 are considered adjacent), and the number of selection methods is denoted as $x_{n}$.
Next, determine the expression for $x_{n}$.
As shown in Figure 1, divide a circular

disk into $n$ sectors, numbered
sequentially as $1,2, \cdots, n$, and
color the sequences $A 、 B 、 C 、 D$ with
different colors.
For any selection

scheme, if the term with index $i$ is from a certain colored sequence, then color the $i$-th sector with that color. Thus, $x_{n}$ becomes 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 $x_{1}=4, x_{2}=12$,
$$
x_{n}+x_{n-1}=4 \times 3^{n-1}(n \geqslant 3) \text {. }
$$

Rewrite equation (1) as
$$
(-1)^{n} x_{n}-(-1)^{n-1} x_{n-1}=-4(-3)^{n-1} \text {. }
$$

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

Adding these equations gives $(-1)^{n} x_{n}=(-3)^{n}+3$.
Thus, $x_{n}=3^{n}+3(-1)^{n}(n \geqslant 2)$.
When $n=13$, $x_{13}=3^{13}-3$, which is the number of card selection methods.