3.3. (Canada, 83). Solve the equation

$$
x!+y!+z!=u!
$$

in natural numbers.
3.3. Let the numbers $x, y, z, u \in \mathbf{N}$ satisfy the original equation, and let $v$ denote the largest of the numbers $x, y, z$. Then $1 \leq v < u$ and

$$
u v! \leq u(u-1)! = u! = x! + y! + z! \leq 3 v!
$$

from which $u v! \leq 3 v!$, hence $u \leq 3$. For $u=3$, we have the equalities

$$
3! = 3 v! = x! + y! + z!
$$

which are valid only in the case $x = y = z = v = 2$. For $u=2$, the equation has no solutions, since in this case

$$
u! = 2 < 3 < x! + y! + z!
$$

Thus, the original equation has the unique solution $x = y = z = 2, u = 3$.