(15) (50 points) Let $p$ be a prime number, $n$ be a positive integer, and $n=n_{0}+n_{1} p+n_{2} p^{2}+\cdots+n_{i} p^{t}$, where $n_{i} \in \mathbf{N}^{*}, 0 \leqslant n_{i} \leqslant p-1, i=0,1,2, \cdots, t$. Let $S_{n}$ denote the set of ordered triples $(a, b, c)$ that satisfy the following conditions:
(1) $a$, $b$, and $c$ are all non-negative integers; (2) $a+b+c=n$; (3) $\frac{n!}{a!b!c!}$ is not divisible by $p$.

How many ordered triples $(a, b, c)$ are there in the set $S_{n}$?
(15) Lemma Let $p$ be a prime, and let $n=n_{0}+n_{1} p+\cdots+n_{t} p^{t}, 0 \leqslant n_{i} \leqslant p-1, 0 \leqslant i \leqslant t$. If $n=e_{0}+e_{1} p+\cdots+e_{t} p^{t}$, where integers $e_{i} \geqslant 0$, $0 \leqslant i \leqslant t$, then
(1) For each $0 \leqslant i \leqslant t$, we have $n_{0}+n_{1}+\cdots+n_{i} \leqslant e_{0}+e_{1}+\cdots+e_{i}$;
(2) If $n_{0}+n_{1}+\cdots+n_{t}=e_{0}+e_{1}+\cdots+e_{t}$, then $n_{i}=e_{i}, 0 \leqslant i \leqslant t$.
Proof of the lemma: We prove (1) and (2) by induction on $n$.
For $n=1$, $n_{0}=1=e_{0}$, so conclusions (1) and (2) hold.
Assume that for $n \leqslant k$, conclusions (1) and (2) hold, consider the case $n=k+1$.
Assume $k+1=n_{0}+n_{1} p+\cdots+n_{t} p^{t}=e_{0}+e_{1} p+\cdots+e_{t} p^{t}$, $0 \leqslant n_{i} \leqslant p-1, e_{i} \geqslant 0, 0 \leqslant i \leqslant t$, then $n_{0} \equiv e_{0}(\bmod p)$.
(i) If $n_{0} \neq 0$, then $1 \leqslant n_{0} \leqslant p-1$ and $e_{0} \geqslant 1$. Therefore,
$$
k=\left(n_{0}-1\right)+n_{1} p+\cdots+n_{t} p^{t}=\left(e_{0}-1\right)+e_{1} p+\cdots+e_{t} p^{t}.
$$

Using the induction hypothesis, we have
$$
\left(n_{0}-1\right)+n_{1}+\cdots+n_{i} \leqslant\left(e_{0}-1\right)+e_{1}+\cdots+e_{i},
$$

which means
$$
n_{0}+n_{1}+\cdots+n_{i} \leqslant e_{0}+e_{1}+\cdots+e_{i}, 0 \leqslant i \leqslant t.
$$

Thus, conclusion (1) holds.
If $n_{0}+n_{1}+\cdots+n_{t}=e_{0}+e_{1}+\cdots+e_{t}$, then
$$
\left(n_{0}-1\right)+n_{1}+\cdots+n_{t} \leqslant\left(e_{0}-1\right)+e_{1}+\cdots+e_{t}.
$$

Using the induction hypothesis, we have $n_{0}-1=e_{0}-1, n_{1}=e_{1}, \cdots, n_{t}=e_{t}$, thus $n_{i}=e_{i}, 0 \leqslant i \leqslant t$.
Conclusion (2) holds.
(ii) If $n_{0}=0$, then $p \mid e_{0}$. Assume $e_{0}=p \cdot e_{0}^{\prime}$, then
$$
\begin{aligned}
\frac{k+1}{p} & =n_{1}+n_{2} p+\cdots+n_{t} p^{t-1} \\
& =\left(e_{0}^{\prime}+e_{1}\right)+e_{2} p+\cdots+e_{t} p^{t-1}.
\end{aligned}
$$

Using the induction hypothesis, we have $n_{1}+\cdots+n_{i} \leqslant\left(\boldsymbol{e}_{0}^{\prime}+\boldsymbol{e}_{1}\right)+e_{2}+\cdots+\boldsymbol{e}_{i}$, thus
$$
\begin{array}{c}
n_{0}+n_{1}+\cdots+n_{i} \leqslant e_{0}^{\prime}+e_{1}+\cdots+e_{i}<e_{0}+e_{1}+\cdots+e_{i}, \\
0 \leqslant i \leqslant t.
\end{array}
$$

Conclusion (1) holds.
If $n_{0}+n_{1}+\cdots+n_{t}=e_{0}+e_{1}+\cdots+e_{t}$, then
$$
n_{1}+n_{2}+\cdots+n_{t}=p \cdot e_{0}^{\prime}+e_{1}+\cdots+e_{t}.
$$

Using the induction hypothesis, for (1), using (1), we have
$$
n_{1}+n_{2}+\cdots+n_{t}=e_{0}^{\prime}+e_{1}+\cdots+e_{t}.
$$

Using (2), we have $p \cdot e_{0}^{\prime} \leqslant e_{0}^{\prime}$, then $e_{0}^{\prime}=0$, thus $n_{0}=e_{0}=0$. Therefore, by (1), we have
$$
\begin{aligned}
n_{1}+n_{2} p+\cdots+n_{1} p^{\imath-1} & =e_{1}+e_{2} p+\cdots+e_{t} p^{\imath-1} \\
& =\frac{k+1}{p} \leqslant k,
\end{aligned}
$$

and
$$
n_{1}+n_{2}+\cdots+n_{t}=e_{1}+e_{2}+\cdots+e_{t}.
$$

Using the induction hypothesis, we have $n_{i}=e_{i}, 1 \leqslant i \leqslant t$. Thus, conclusion (2) holds.
In summary, conclusions (1) and (2) both hold.
Proof of the original proposition: Let $p$ be a prime, and let $n$ be a positive integer. If $p^{a} \mid n$, but $p^{a+1} \nmid n$, then we denote $V_{p}(n)=a$, thus
$$
V_{p}(n!)=\sum_{k=1}^{+\infty}\left[\frac{n}{p^{k}}\right].
$$

If $n=n_{0}+n_{1} p+\cdots+n_{\iota} p^{t}, 0 \leqslant n_{i} \leqslant p-1, 0 \leqslant i \leqslant t$, then
$$
V_{p}(n!)=\frac{n-\left(n_{0}+n_{1}+\cdots+n_{t}\right)}{p-1}.
$$

Assume
$$
\left\{\begin{array}{ll}
n=n_{0}+n_{1} p+\cdots+n_{1} p^{2}, & 0 \leqslant n_{i} \leqslant p-1, \\
a=a_{0}+a_{1} p+\cdots+a_{t} p^{2}, & 0 \leqslant a_{i} \leqslant p-1, \\
b=b_{0}+b_{1} p+\cdots+b_{t} p^{2}, & 0 \leqslant b_{i} \leqslant p-1, \\
c=c_{0}+c_{1} p+\cdots+c_{t} p^{2}, & 0 \leqslant c_{i} \leqslant p-1,
\end{array}\right.
$$

where $n=a+b+c$, then $p \nmid \frac{n!}{a!b!c!}$ if and only if
$$
V_{p}(n!)=V_{p}(a!)+V_{p}(b!)+V_{p}(c!).
$$

Using (3), we know that $p \nmid \frac{n!}{a!b!c!}$ if and only if $\sum_{i=0}^{t} n_{i}=\sum_{i=0}^{t}\left(a_{i}+b_{i}+c_{i}\right)$.
Using lemma (1) and (2), we have $p \nmid \frac{n!}{a!b!c!}$ if and only if $n_{i}=a_{i}+b_{i}+c_{i}$, $0 \leqslant i \leqslant t$.
Therefore, the number of ordered triples $(a, b, c)$ that satisfy the condition is
$$
\left|S_{n}\right|=\mathrm{C}_{n_{0}+2}^{2} \cdot \mathrm{C}_{n_{1}+2}^{2} \cdots \cdot \mathrm{C}_{n_{t}+2}^{2}.
$$