10. (2005 Vietnam Mathematical Olympiad) Find all natural number triples $(x, y, n)$ that satisfy the equation $\frac{x!+y!}{n!}=3^{n}$ (with the convention that $0!=1$).
10. Solution: Suppose $(x, y, n)$ is a triplet of natural numbers satisfying
\[
\frac{x!+y!}{n!}=3^{n}
\]
It is easy to see that $n \geqslant 1$.
Assume $x \leqslant y$, consider the following two cases:
(1) $x \leqslant n$, it is easy to see that $\frac{x!+y!}{n!}=3^{n} \Leftrightarrow 1+\frac{y!}{x!}=3^{n} \cdot \frac{n!}{x!}$
From equation (2), we can deduce $1+\frac{y!}{x!} \equiv 0(\bmod 3)$.
Since the product of three consecutive integers is divisible by 3, and because $n \geqslant 1$, we have $x < n$, which leads to a contradiction. Therefore, this case is not possible.
(2) $x > n$, it is easy to see that $\frac{x!+y!}{n!}=3^{n} \Leftrightarrow \frac{x!}{n!}\left(1+\frac{y!}{x!}\right)=3^{n}$.
Since $n+1$ and $n+2$ cannot both be multiples of 3, from equation (7) we can deduce $x=n+1$.
Thus, equation (7) becomes $n+1+\frac{y!}{n!}=3^{n}$.
Since $y \geqslant x$, then $y \geqslant n+1$.
Let $A=\frac{y!}{(n+1)!}$, substituting into equation (8) gives $(n+1)(A+1)=3^{n}$.
If $y \geqslant n+4$, then $A \equiv 0(\bmod 3)$, hence $A+1$ cannot be a multiple of 3.
Therefore, from equation (9) we get $y \leqslant n+3$, so $n+1 \leqslant y \leqslant n+3$.
(i) If $y=n+3$, then $A=(n+2)(n+3)$,
From equation (9) we get $(n+1)[1+(n+2)(n+3)]=3^{n}$, i.e., $(n+2)^{3}-1=3^{n}$
This implies $n>2, n+2 \equiv 1(\bmod 3)$.
Let $n+2=3 k+1, k \geqslant 2$, from equation (10) we have
$9 k\left(3 k^{2}+3 k+1\right)=3^{3 k-1}$, which contradicts $\left(3 k^{2}+3 k+1,3\right)=1$, so $y \neq n+3$.
(ii) If $y=n+2$, then $A=n+2$.
From equation (9) we get $(n+1)(n+3)=3^{n}$.
For $n \geqslant 1$, $(n+1)$ and $(n+3)$ cannot both be multiples of 3, so $y \neq n+2$.
(iii) If $y=n+1$, then $A=1$. From equation (9) we get $2(n+1)=3^{n}$.
Clearly, there is no $n$ that satisfies this equation, so $y \neq n+1$.
In summary, the triplets $(x, y, n)$ that satisfy equation (1) are $(0,2,1)$, $(2,0,1)$, $(1,2,1)$, or $(2,1,1)$.
Upon verification, these four solutions satisfy the condition.