Find all positive integers $a$, $b$, $c$ such that $a$, $b$, $c$ satisfy $(a!)(b!)=a!+b!+c!$.
(2002-2003 British Mathematical Olympiad)
Let $\alpha(n)$ be the exponent of 2 in the prime factorization of $n$.
If $a \neq b$, assume without loss of generality that $a > b$.
Clearly, $b \neq 1$ or 2.
Thus, $c! = (a!)(b!) - a! - b!$
$> (a!)(b! - 2) > a!$.
If $a = b + 1$, then
$(b + 1)! = b + 2 + \prod_{t=b+1}^{c} t$.
The left side of the above equation is a multiple of $b + 1$, while the right side is not a multiple of $b + 1$, leading to a contradiction. Therefore, $a \geqslant b + 2$. Then,
$\alpha(a! + b! + c!) = \alpha(b!) < \alpha((a!)(b!))$, which is a contradiction. This is impossible.
Thus, $a = b$.
Therefore, $(a!)(a! - 2) = c!$.
Clearly, $a \geqslant 3, 3 \times (a! - 2) = \prod_{t=a+1}^{c} t$.
Hence, $c = a + 1$ or $c = a + 2$.
Upon inspection, the only solution $(a, b, c) = (3, 3, 4)$ is found when $c = a + 1$, which is the unique solution.