Example 3.17 gives the array $\mathrm{I}$:
$$
\begin{array}{cccccccccc} 
& 1 & 2 & 3 & 4 & \cdots & n-3 & n-2 & n-1 & \\
1 & 2 & 3 & 4 & 5 & \cdots & n-2 & n-1 & n \\
2 & 3 & 4 & 5 & 6 & \cdots & n-1 & n & & (n \geqslant 2),
\end{array}
$$

Select a number from the $k$-th column $(k=1,2, \cdots, n)$, denoted as $a_{k}$. If $a_{1}, a_{2}, \cdots, a_{n}$ are distinct, then the ordered $n$-tuple $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ is called a system of distinct representatives (SDR) of the array $\mathrm{I}$, and $a_{k}$ is called the representative of the $k$-th column. Let $u_{n}$ denote the number of SDRs of the array $\mathrm{I}$. Find $u_{n}$.
Solution: It is easy to see that $u_{2}=2, u_{3}=3$. When $n \geqslant 4$, the $u_{n}$ SDRs of array $\mathrm{I}$ can be divided into the following two categories:
(1) SDRs where the representative of the $n$-th column is $n$.
The number of SDRs in this category is equal to the number of SDRs of the array
$$
\begin{array}{llllllll} 
& 1 & 2 & 3 & 4 & \cdots & n-3 & n-2 \\
1 & 2 & 3 & 4 & 5 & \cdots & n-2 & n-1 \\
2 & 3 & 4 & 5 & 6 & \cdots & n-1 &
\end{array}
$$

which is $u_{n-1}$.
(2) SDRs where the representative of the $n$-th column is $n-1$.
For SDRs in this category, the representative of the $(n-1)$-th column must be $n$, so the number of SDRs in this category is equal to the number of SDRs of the array
$$
\begin{array}{llllllll} 
& 1 & 2 & 3 & 4 & \cdots & n-4 & n-3 \\
1 & 2 & 3 & 4 & 5 & \cdots & n-3 & n-2 \\
2 & 3 & 4 & 5 & 6 & \cdots & n-2 &
\end{array}
$$

which is $u_{n-2}$.
By the principle of addition, we have
$$
u_{n}=u_{n-1}+u_{n-2} \quad(n \geqslant 4) .
$$

Let $u_{0}=1, u_{1}=1$, then the above formula holds for $n \geqslant 2$. Therefore, $u_{n}$ is the Fibonacci number $f(n)$, thus
$$
u_{n}=\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{n+1} .
$$