7. Given a three-digit number $x y z(1 \leqslant x \leqslant 9,0 \leqslant y, z$ $\leqslant 9)$. If $x y z=x!+y!+z!$, then the value of $x+y+z$ is
7. 10 .

Since $6!=720$, we have $0 \leqslant x, y, z \leqslant 5$.
And $1!=1, 2!=2, 3!=6, 4!=24, 5!=120$.
By observation, we get $x=1, y=4, z=5$. Therefore,
$$
x+y+z=10 \text {. }
$$