10. (15 points) For a positive integer $n$, denote
$$
n!=1 \times 2 \times \cdots \times n \text {. }
$$

Find all positive integer tuples $(a, b, c, d, e, f)$ such that
$$
a!=b!+c!+d!+e!+f!,
$$

and $\quad a>b \geqslant c \geqslant d \geqslant e \geqslant f$.
10. From the problem, we know that $a! \geqslant 5 \cdot f! \geqslant 5$. Therefore, $a \geqslant 3$.

From $a > b$, we get $a! \geqslant a \cdot b!$.
Combining this with the problem statement, we have $a \cdot b! \leqslant a! \leqslant 5 \cdot b!$.
Thus, $a \leqslant 5$. Therefore, $3 \leqslant a \leqslant 5$.
When $a=3$, $b<3$. In this case,
$b! + c! + d! + e! + f! = 6$.
When $b=2$, $c! + d! + e! + f! = 4$.
Hence, $c=d=e=f=1$.
When $b=1$, $c=d=e=f=1$, which does not satisfy the problem statement.
When $a=4$, $b<4$, in this case,
$b! + c! + d! + e! + f! = 24$.
When $b=3$, $c! + d! + e! + f! = 18$.
Since $4 \times 2! < 18$, we have
$$
c=3, d! + e! + f! = 12;
$$

Since $3 \times 2! < 12$, we have $d=3, e! + f! = 6$, which has no solution.
When $b=2$,
$b! + c! + d! + e! + f! \leqslant 5 \times 2! < 24$,

which has no solution.
When $a=5$, $b \leqslant 4$, in this case,
$b! + c! + d! + e! + f! = 120$.
Also, $120 = b! + c! + d! + e! + f!$
$\leqslant 5 \cdot b! \leqslant 5 \times 4! = 120$,
thus $b=c=d=e=f=4$.
In summary, the solutions that satisfy the problem statement are
$$
\begin{array}{l}
(a, b, c, d, e, f) \\
=(3,2,1,1,1,1),(5,4,4,4,4,4) .
\end{array}
$$