$$\begin{array}{l} 22. \text{Let } x \leqslant y \leqslant z, \text{ and } w! = z! + y! + x! \\ z \leqslant 2 \end{array}$$

If $z=1$, then $x=y=z=1$, at this time $w!=3$, there is no such $w$, so $z=2$. At this time $w \geqslant 3$, hence
$$w!\equiv 0(\bmod 3)$$

Therefore,
$$x!+y!\equiv 1(\bmod 3)$$

And
thus it can only be
$$\begin{array}{l} x \leqslant y \leqslant 2 \\ x=y=2 \end{array}$$

At this time
hence
$$(x, y, z, w)=(2,2,2,3)$$