Example 1 (2007 Tianjin College Entrance Examination Question) Use 6 different colors to color the 4 squares in the figure, with each square being one color, requiring the use of at most 3 colors and adjacent squares must not be the same color. The number of different coloring methods is $\qquad$ (answer with a number).
Solution 1 (Interval Position Classification Discussion Method) Since positions 1 and 2 are painted in different colors, and positions 1 and 3 can be painted in the same color or different colors, and once the coloring situation of positions 1 and 3 is determined, the coloring situation of positions 2 and 4 is also determined, so we classify and discuss according to the coloring situation of positions 1 and 3.

When positions 1 and 3 are painted in the same color, there are $\mathrm{C}_{6}^{1}$ ways to paint positions 1 and 3, and there are $\mathrm{C}_{5}^{1}$ ways to paint positions 2 and 4, so there are $\mathrm{C}_{6}^{1} \mathrm{C}_{5}^{1} \mathrm{C}_{5}^{1} = 150$ ways to paint in this situation.

When positions 1 and 3 are painted in different colors, there are $\mathrm{A}_{6}^{2}$ ways to paint positions 1 and 3, 2 ways to paint position 2, and since at most 3 colors are used, there are 2 ways to paint position 4, so there are $\mathrm{A}_{6}^{2} \times 4 \times 2 = 240$ ways to paint in this situation.

Therefore, there are a total of $150 + 240 = 390$ ways to paint that meet the conditions.

Comment: In coloring problems, since adjacent positions are not painted in the same color, and interval positions can be painted in the same color or different colors, and once the coloring situation of the interval positions is determined, the coloring situation of other positions is also determined, the interval position classification discussion method is a general method to solve coloring problems.

Solution 2 (Color Type Classification Discussion Method) Since 4 positions can be painted in 2 colors or 3 colors, we can classify and discuss according to the type of colors used.

When painting with 2 colors, positions 1 and 3 are the same color, and positions 2 and 4 are the same color, in this situation there are $\mathrm{A}_{6}^{2} = 30$ ways to paint;

When painting with 3 colors, there must be two positions of the same color, and the same color positions can be 1,3 or 1,4 or 2,4, in this situation there are $3 A_{6}^{3} = 3 \times 6 \times 5 \times 4 = 360$ ways to paint.

Therefore, there are a total of $30 + 360 = 390$ ways to paint that meet the conditions.

Comment: When the colors available for coloring are optional, classifying by the type of colors helps determine the coloring of each position, promoting the solution of the problem.

Solution 3 (Exclusion Method) Since 4 positions can be painted in at most 4 colors, assuming there is no limit on the number of colors used, there are $6 \times 5 \times 5 \times 5 = 750$ ways to paint in the order of 1, 2, 3, 4, but there are $\mathrm{A}_{6}^{4} = 360$ ways to paint 4 colors that do not meet the conditions, so there are $750 - 360 = 390$ ways to paint that meet the conditions.

Comment: In coloring problems, when the opposite situation is simple, using the permutation method helps simplify and solve complex and difficult problems, which is a wise choice.