2. Solve the equation in natural numbers $x, y, z$ (where $x!$ denotes the factorial of $x$)

$$
\frac{x+y}{z}=\frac{x!+y!}{z!}
$$
Solution. Answer: $(a, b, c)$, where each of $a, b, c$ is independently 1 or $2 ;(a, a, a)$ for $a \geqslant 3$.

Rewrite the equation as $(x+y)(z-1)!=x!+y!$ !

1) Let $z>x$ and $z>y$. Then $(x+y)(z-1)!=x!+y!\leqslant 2(z-1)!$ ! From this, $x+y \leqslant 2$ and $x=y=1$. In this case, $(z-1)!=1$ and $z=1$ or $z=2$. Thus, in the considered case, we get the triplet $(1,1,2)$.
2) Let $z=y$. Then the equation takes the form $(x+z)(z-1)!=x!+z$ ! or $(z-$ $1)!=(x-1)!$. Then either $x=y=z$, or one of the two: $x=1, z=2 ; x=2, z=$ 1. We get the triplets $(a, a, a), a \in \mathbb{N},(2,1,1),(1,2,2)$. Similarly, in the case $z=x$, we additionally get $(1,2,1)$ and $(2,1,2)$.
3) Let $x>z$ and $y>z$. Then $x(z-1)!\leqslant x!$ and $y(z-1)!\leqslant y!$, and equality holds only if $z=1$. Then $x=y=2$ and we add another solution $(2,2,1)$.