4. As shown in the figure: The number of ways to plant in $A$ and $B$ is $P_{6}^{2}$. If $C$ and $B$ are planted with the same type of plant, then the number of ways to plant in $C$ and $D$ is $C_{5}^{1}$. If $C$ and $B$ are planted with different types of plants, then the number of ways to plant in $C$ and $D$ is $C_{4}^{1} C_{4}^{1}$. Therefore, the number of ways to plant in $C$ and $D$ is $C_{5}^{1} + C_{4}^{1} C_{4}^{1}$. Similarly, the number of ways to plant in $E$ and $F$ is $C_{5}^{1} + C_{4}^{1} C_{4}^{1}$. Therefore, the total number of different planting methods that meet the requirements is $P_{6}^{2}\left(C_{5}^{1} + C_{4}^{1} C_{4}^{1}\right)^{2} = 13230$.
$4.2 \times 3$ rectangular flower bed is divided into 6 $1 \times 1$ small square areas: $A, B, C, D, E, F$. In each area, a type of plant is planted, and the plants in adjacent areas must be different. Now, there are 6 types of plants to choose from. How many different planting schemes are there?
(2002 Hunan Province Middle School Mathematics Summer Camp Question)