Example 4 Let $p$ be a prime, $n \in Z_{+}$, and
$$
n=n_{0}+n_{1} p+\cdots+n_{t} p^{t},
$$

where, $n_{i} \in \mathbf{N}, 0 \leqslant n_{i} \leqslant p-1, i=0,1, \cdots, t$.
Let $S_{n}$ denote the set of ordered triples $(a, b, c)$ that satisfy the following conditions:
(1) $a, b, c \in \mathbf{N}$;
(2) $a+b+c=n$;
(3) $p \uparrow \frac{n!}{a!\cdot b!\cdot c!}$.
Question: How many ordered triples $(a, b, c)$ are there in the set $S_{n}$?
[3]
(2012, National High School Mathematics League Jiangsu Province Re-test)
【Analysis】By the lemma, imitating the proof of Corollary 1, we have
$$
\begin{array}{l}
v_{p}\left(\frac{n!}{a!\cdot b!\cdot c!}\right)=0 \\
\Leftrightarrow \tau_{p}(a)+\tau_{p}(b)+\tau_{p}(c)-\tau_{p}(n)=0
\end{array}
$$
$\Leftrightarrow a, b, c$ do not produce a carry in any digit when performing column addition in $p$-ary representation.
$$
\begin{array}{l}
\text { Let } a=\sum_{i=0}^{t} a_{i} p^{i}, b=\sum_{i=0}^{t} b_{i} p^{i}, c=\sum_{i=0}^{t} c_{i} p^{i} \\
\Rightarrow a_{i}+b_{i}+c_{i}=n_{i}(i=0,1, \cdots, t) .
\end{array}
$$

Note that, the indeterminate equation $x+y+z=m$ has $\mathrm{C}_{m+2}^{2}$ sets of non-negative integer solutions.
Therefore, the number of ordered triples satisfying the condition is
$$
\left|S_{n}\right|=\prod_{i=0}^{t} \mathrm{C}_{n_{i}+2}^{2} .
$$