Example 6 As shown in Figure 22, use 6 different colors to color the 4 different regions $A, B, C, D$, requiring that adjacent regions be colored with different colors. How many different coloring methods are there?
Solution 1: Complete in four steps according to the order of $A-B-C-D$.
(1) Color region $A$ with no restrictions, 6 colors can be used, giving 6 ways to color.
(2) Color region $B$, which must be a different color from $A$, only 5 remaining colors can be used, giving 5 ways to color.
(3) Color region $C$, which must be a different color from both $A$ and $B$, only 4 remaining colors can be used, giving 4 ways to color.
(4) Color region $D$, which only needs to be a different color from $C$, hence there are 5 ways to color.
Multiplying gives $6 \times 5 \times 4 \times 5=600$ different ways to color.
Solution 2: Complete in steps according to the order of $A-D-C-B$, coloring $A$ has 6 methods; coloring $D$ has 6 methods; coloring $C$ must be different from $A$ and $D$, hence there are 4 methods; coloring $B$ must be different from $A$ and $C$, hence there are 4 methods.
Multiplying gives $6 \times 6 \times 4 \times 4=576$ different ways to color.
Solution 3: Complete in steps according to the order of $A-D-B-C$, coloring $A$ has 6 methods; coloring $D$ has 6 methods; coloring $B$ must be different from $A$, hence there are 5 methods; coloring $C$ must be different from $A$, $B$, and $D$, hence there are 3 methods.
Multiplying gives $6 \times 6 \times 5 \times 3=540$ different ways to color.
Analysis: All three solutions use the step-by-step calculation (multiplication) method, but due to different orders, the answers are different, and there must be repetition or omission.

Reviewing Solution 1, no problems are found. Reflecting on Solution 2, we can see that regions $A$ and $D$ are not adjacent, allowing the same color (hence both have 6 ways to color). When $A$ and $D$ are the same color, $C$ is not 4 ways to color, but 5 ways to color, resulting in $6 \times 5 \times 4 = 120$ ways to color. However, adding 576 and 120 would result in repetition. The correct approach is to consider the cases where $A$ and $D$ are different colors and the same color, then add them, resulting in
$6 \times 5 \times 4 \times 4 + 6 \times 5 \times 4 = 600$.
Similarly, in Solution 3, there is also an omission of the coloring of region $C$ when $A$ and $D$ are the same color or $B$ and $D$ are the same color. The correct solution can be divided into 3 cases:
$A$ and $D$ are the same color and $B$ and $D$ are different colors, or $A$ and $D$ are different colors and $B$ and $D$ are the same color, which is equivalent to 3 regions each being colored 1 color, each having $6 \times 5 \times 4 = 120$ ways to color; $A$ and $D$ are different colors and $B$ and $D$ are different colors, which is equivalent to 4 regions each being colored 1 color, having $6 \times 5 \times 4 \times 3 = 360$ ways to color.
In total, $120 + 120 + 360 = 600$ different ways to color.
To reduce errors, the following two suggestions are proposed:
(1) Correspond regions to points, and adjacent relationships to lines between two points. This can simplify the graph and highlight the relationships.
(2) Each time coloring starts from the point with the most adjacent regions. This can better avoid repetition and omission.

Solution 4: Correspond regions $A$, $B$, $C$, $D$ to points $A$, $B$, $C$, $D$, and draw a line between two points when two regions are adjacent, resulting in a graph (Figure 23). The problem is transformed into point coloring, and coloring is done in steps according to the number of lines, in the order of $C-A-B-D$. Point $C$ has 6 ways to color, point $A$ has 5 ways to color, point $B$ has 4 ways to color, and point $D$ has 5 ways to color.
Multiplying gives $6 \times 5 \times 4 \times 5 = 600$ different ways to color.
Coloring in the order of $C-B-A-D$ or $C-A-D-B$ has the same effect. Additionally, the problem can also be solved by classifying based on the use of colors.