Example 12 (2001 National High School Competition Question) In the six regions of a regular hexagon, ornamental plants are to be planted (as shown in Figure 12-2). The requirement is that the same type of plant is planted in the same region, and different types of plants are planted in adjacent regions. Given 4 different types of plants to choose from, how many planting schemes are there?
Now, let's solve this problem using probability knowledge.
Let's assume the boundary of $A F$ is unfolded as shown in Figure 12-3.
There are 4 different types of plants to choose from, denoted as $a, b, c, d$. According to the problem, the plants in adjacent plots from $A$ to $F$ must be different, and the plants in the first and last plots, $A$ and $F$, must also be different.
Considering that there are 4 choices for the plant in the $A$ plot, let's assume it is $a$. Then, the $B$ plot has 3 choices, let's say $b$; the $C$ plot can choose from $a, c, d$, and so on. By the time we reach the $E$ plot, there are still 3 choices. However, the choice for the $F$ plot depends on whether the plants in the $A$ and $E$ plots are the same. If the plants in the $A$ and $E$ plots are the same, say $a$, then the $F$ plot can choose from $b, c, d$, which gives $C_{3}^{1}$ ways; if the plants in the $A$ and $E$ plots are different, say $a, b$, then the $F$ plot can only choose from $c, d$, which gives $C_{2}^{1}$ ways.

Next, let's examine the probability that the plants in the $A$ and $E$ plots are the same. According to the principle that the sum of the probabilities of complementary events is 1, and the probability of two independent events occurring simultaneously is the product of the probabilities of each event. For this problem, if the plant in the $A$ plot is $a$, then the probability of the $B$ plot being $a$ is $0$. The probability of the $C$ plot being $a$ is $\frac{1}{3}$. According to the problem's restrictions, the probability of the $D$ plot being $a$ is $\left(1-\frac{1}{3}\right) \times \frac{1}{3}=\frac{2}{9}$, and so on, the probability of the $E$ plot being $a$ is $\left(1-\frac{2}{9}\right) \times \frac{1}{3}=\frac{7}{27}$. Correspondingly, the probability of the plants in the $E$ and $A$ plots being different is $1-\frac{7}{27}=\frac{20}{27}$. Therefore, by the two principles, the total number of ways is $4 \times 3 \times 3 \times 3 \times 3 \times\left(3 \times \frac{7}{27}+2 \times \frac{20}{27}\right)=12 \times 61=732$ (ways).