Example 6 For a positive integer $n$, denote $n!=1 \times 2 \times \cdots \times n$. Find all positive integer tuples $(a, b, c, d, e, f)$, such that
$$
\begin{array}{l}
\quad a!=b!+c!+d!+e!+f!, \\
\text { and } \quad a>b \geqslant c \geqslant d \geqslant e \geqslant f .[1]
\end{array}
$$
Solve: From equation (1), we get
$$
a!>5(f!) \geqslant 5 \Rightarrow a \geqslant 3 \text {. }
$$

From $a>b \Rightarrow a!\geqslant a(b!)$.
Combining $a!=b!+c!+d!+e!+f!\leqslant 5 b!$, we get $a \leqslant 5$.
When $a=3$, it is not difficult to determine the positive integer tuple as $(3,2,1,1,1,1)$.
When $a=4$,
(1) If $b=3$, then $c!+d!+e!+f!=18$. From $4(2!)<18$, we get $c=3$.
Thus, $d!+e!+f!=12$.
From $3(2!)<12$, we get $d=3$.
Therefore, $e!+f!=6$, which clearly has no solution.
(2) If $b \leqslant 2$, from
$$
b!+c!+d!+e!+f!\leqslant 5(2!),
$$

we know there is no solution in this case.
When $a=5$,
$$
\begin{array}{l}
120=b!+c!+d!+e!+f! \\
\leqslant 5(b!) \leqslant 5(4!)=120 .
\end{array}
$$

Thus, $b=c=d=e=f=4$.
Therefore, the positive integer tuple is $(5,4,4,4,4,4)$.
In summary, the positive integer tuples that satisfy the problem are
$$
(3,2,1,1,1,1),(5,4,4,4,4,4) \text {. }
$$