881*. Solve the equation in natural numbers

$$
x!+y!=z!
$$
$\triangle$ Consider two cases.

1) Let $x<y$. Then the equation can be reduced by $x$!:

$$
1+(x+1)(x+2) \ldots y=(x+1)(x+2) \ldots z
$$

From here, 1 is divisible by $x+1$. But this is impossible, since the sum $x+1$ is at least 2.

2) Let $x=y$. We get:

$$
x!+x!=z!, \quad 2 x!=z!
$$

This equation can also be reduced by $x$!:

$$
2=(x+1)(x+2) \ldots z
$$

The last equality is possible only if there is only one factor on the right $-x+1=2$. Hence, $x=1$. We find $z$

$$
z!=2 x!=2 \cdot 1!=2, \quad z=2
$$

Answer: $(1 ; 1 ; 2)$.