Promotion 2 Select $n$ $(1 \leqslant n \leqslant 10)$ digits $a_{1}, a_{2}, \cdots, a_{n}$ from the ten digits $0,1, \cdots, 9$, to form an $n$-digit number $P$. Sort all $P$ in ascending order, find what $P_{m}$ equals to.
Solution: As defined above.
Let $b_{1}, b_{2}, \cdots, b_{n}$ be a permutation of $a_{1}, a_{2}, \cdots, a_{n}$ in ascending order. Now, we analyze the numbers formed by $b_{i}(i=1,2, \cdots, n)$.

Construct a mapping $f$: replace each $b_{i}$ in every number $P$ with $i-1$, forming a new number $Q$ (for example, $\left.f\left(b_{3} b_{2} b_{4}\right)=213\right)$.
Similarly, we can obtain the following conclusions:
(1) $Q$ is composed of the digits $0,1, \cdots, n-1$;
(2) Due to the ordered nature of $\left\{b_{i}\right\}$, if $b_{\alpha}<b_{\beta}$, then $\alpha-1<\beta-1$, so $f$ is monotonic.

Therefore, sorting $Q$ in ascending order, we must have $f\left(P_{i}\right) = Q_{i}$. Thus, the set of numbers formed by $Q$ also meets the requirement of being consecutive non-negative integers in base $n$, and they are one-digit, two-digit, $\cdots$, $n$-digit numbers. Hence,
\[
\begin{array}{l}
Q_{m}=\varphi_{n}(m-1), \\
P_{m}=f^{-1}\left(Q_{m}\right)=f^{-1}\left(\varphi_{n}(m-1)\right) .
\end{array}
\]