Example 7 (2003 British Mathematical Olympiad) Find all positive integers $a, b, c$ satisfying $(a!) \cdot(b!)=a! + b! + c!$

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Without loss of generality, assume $a \geqslant b$, then the original equation becomes $a!=\frac{a!}{b!}+1+\frac{c!}{b!}$.
Since the three terms $a!, \frac{a!}{b!}, 1$ in the above equation are integers, it follows that $c \geqslant b$.
Since each term on the right side is a positive integer, their sum is at least 3. Therefore, $a \geqslant 3$, and $a!$ is even. Thus, $\frac{a!}{b!}$ and $\frac{c!}{b!}$ have exactly one term that is odd.
(1) Assume $\frac{a!}{b!}$ is odd, then either $a=b$, or $\frac{a!}{b!}=b+1$ and $b+1$ is odd, $a=b+1$.
(i) If $a=b$, then $a!=2+\frac{c!}{a!}$,
When $a=3$, we have $b=3, c=4$,
When $a>3$, since $a!-2 \equiv-2(\bmod 3)$, so
$c=a+1$ or $a+2$ (otherwise $\frac{c!}{a!} \equiv 0(\bmod 3)$).
$\frac{c!}{a!}=a+1$ or $(a+1)(a+2)$,
$a!=(a+3)$ or $(a+1)(a+2)+2$,
When $a=4$ or 5, it does not satisfy (1).
When $a \geqslant 6$, it is clear that the left side of (1) is greater than the right side, so the original equation has no solution.
(ii) If $a=b+1$, where $b$ is even,

then the original equation becomes $(b+1)!=b+2+\frac{c!}{b!}$,
Taking both sides modulo $b+1$ gives $\frac{c!}{b!} \equiv-1(\bmod (b+1))$.
Since $\frac{a!}{b!}$ is odd, $\frac{c!}{b!}$ is even, hence $c>b$.
Thus, $\frac{c!}{b!} \equiv 0(\bmod (b+1))$, which is a contradiction.
(2) Assume $\frac{a!}{b!}$ is even, $\frac{c!}{b!}$ is odd, then $c=b$ or $c=b+1$ ($b$ is even).
(i) If $c=b$, then the equation becomes $(a!) \cdot(b!)=(a!)+2 \times(b!)$,
So $\frac{a!}{b!}(b!-1)=2$. Hence $\frac{a!}{b!}=2, b!-1=1$.
Thus $b=2, a!=4$, which is a contradiction.
(ii) If $c=b+1$, then the equation becomes $(a!)(b!)=(a!)+(b+2)(b!)$.

So $(a!)(b!-1)=(b+2)(b!)$.
Since $(b!, b!-1)=1$, it follows that $(b!-1) \mid b+2$.
Since $b$ is even, so $b=2, a!=8$, which is impossible.
In conclusion, the original equation has a unique solution $a=3, b=3, c=4$.