【Example 3】 $n$ boys and $n$ girls stand in a circle to dance, requiring that no two boys are adjacent, and no two girls are adjacent. How many different ways are there to arrange them?
Solution 1: First, let the boys form a circle, which has $(n-1)!$ ways of standing. There is a gap between each two boys, making a total of $n$ gaps. Let the $n$ girls arrange themselves in these $n$ gaps, which has $n!$ ways of arranging. Therefore, there are a total of $(n-1)! \cdot n!$ different ways of standing. Note that after the boys are arranged, the $n$ gaps become $n$ distinct positions, so the girls' standing method should be considered as a permutation, and should not be treated as a circular arrangement again.

Solution 2: First, let the boys form a line, which has $n!$ ways of arranging. Then, choose $n$ adjacent gaps from the $n+1$ gaps (there is a gap between each two people, and the ends are also counted as gaps), which has 2 ways of choosing (Figure 8). The $n$ girls still need to arrange themselves in the chosen $n$ gaps, which has $n!$ ways of arranging. Therefore, there are a total of $2 \cdot (n!)^2$ ways of standing in a line. Then, by the relationship between the number of permutations and the number of circular arrangements, we know that when they stand in a circle, there are
$$
\frac{2(n!)^2}{2n} = n! \cdot (n-1)!
$$
ways of circular arrangement.