For example, the five-digit numbers formed by the digits $1, 2, 3, 4, 5$ with repetition, arranged in ascending order. Ask:
(1) What are the positions of 22435 and 43512?
(2) What is the 200th number? ${ }^{[1]}$
Analysis: Let the $n$-th number after sorting be denoted as $a_{n}$. Since the numbers do not contain 0, we cannot directly use the quinary system. Therefore, subtract 11111 from all $a_{n}$, and denote the result as $b_{n}$. Thus, we can draw the following conclusions:
(i) Each number in $b_{n}$ consists of some of the digits $0, 1, 2, 3, 4$;
(ii) The $i$-th number in $\left\{a_{n}\right\}$, denoted as $a_{i}$, when subtracted by 11111, becomes $b_{i}$, which is the $i$-th number in $\left\{b_{n}\right\}$, meaning the order of the sequence remains unchanged after the transformation;
(iii) $b_{n}$ is a non-negative integer with one, two, three, four, or five digits.

Thus, $\left\{b_{n}\right\}$ meets the requirement of being a sequence of consecutive quinary integers, and the method in the problem can be used to solve it.
Solution: (1) Note that
$$
\begin{array}{l}
22435-11111=11324, \\
43512-11111=32401 .
\end{array}
$$

Therefore, the order numbers of 22435 and 43512 are
$$
\begin{array}{l}
(11324)_{5}+1=840, \\
(32401)_{5}+1=2227 .
\end{array}
$$
(2) Since $200=(1244)_{5}+1$, the 200th

number is $1244+11111=12355$.