Example 2 If $A_{1}, A_{2}, \cdots, A_{k}$ is a cover of $X$, and each element of $X$ belongs to exactly two of the sets $A_{1}, A_{2}, \cdots, A_{k}$, then $A_{1}, A_{2}, \cdots, A_{k}$ is called a double cover of $X$. Find the number of double covers for $k=3$.
Solve for each element in $X$ belonging to two of $A_{1}, A_{2}, A_{3}$, hence there are three possibilities. The assignment of $n$ elements has a total of $3^{n}$ possibilities. Excluding the three cases where exactly one of $A_{1}, A_{2}, A_{3}$ is an empty set, there are $3^{n}-3$ possibilities. Since the order of $A_{1}, A_{2}, A_{3}$ is not considered, the number of double covers for $k=3$ is $\frac{3^{n}-3}{3!}$.
Note: Let the number of double covers formed by $k$ sets be $a_{k}$, then
$$
a_{k}=\frac{1}{k!}\left(\left(C_{k}^{2}\right)^{n}-k a_{k-1}\right) .
$$