10. (2002-2003 British Mathematical Olympiad) Find all positive integers $a, b, c$, such that $(a!)(b!)= a! + b! + c!$.
10. Let's assume $a \geqslant b$, the original equation can be transformed into $a!=\frac{a!}{b!}+1+\frac{c!}{b!}$. Since $a!, \frac{a!}{b!}, 1$ are all integers, it follows that $c \geqslant b$, and $a!=\frac{a!}{b!}+1+\frac{c!}{b!} \geqslant 3$, hence $a \geqslant 3$, and thus $a!$ is even. Therefore, among $\frac{a!}{b!}$ and $\frac{c!}{b!}$, exactly one is odd.
(1) Assume $\frac{a!}{b!}$ is odd, then either $a=b$ or $\frac{a!}{b!}=b+1$ and $b+1$ is odd, with $a=b+1$.
(i) If $a=b$, then $a!=2+\frac{c!}{a!}$. When $a=3$, we have $b=3, c=4$. When $a \geqslant 4$, since $a!-2$ is not divisible by 3, $c=a+1$ or $c=a+2$, and $\frac{c!}{a!}=a+1$ or $(a+1)(a+2)$, thus $a!=a+3$ or $a!=(a+1)(a+2)+2$. When $a=4$ or 5, the above equations do not hold. When $a \geqslant 6$, $a!>(a+1)(a+2)+2>a+3$, so the original equation has no solution.
(ii) If $a=b+1$, where $b$ is even, then the original equation becomes $(b+1)!=b+2+\frac{c!}{b!}$. Since $\frac{a!}{b!}$ is odd, $\frac{c!}{b!}$ is even, hence $c \geqslant b+1$, and $\frac{c!}{b!}$ is divisible by $b+1$, thus $b+2$ is divisible by $b+1$, which is a contradiction.
(2) Assume $\frac{a!}{b!}$ is even, and $\frac{c!}{b!}$ is odd, then $c=b$ or $c=b+1$ (where $b$ is even).
(i) If $c=b$, then the original equation becomes $(a!)(b!)=a!+2(b!)$, $\frac{a!}{b!}(b!-1)=2$, hence $\frac{a!}{b!}=2, b!-1=1$, so $b=2, a!=4$ is impossible.
(ii) If $c=b+1$, then the original equation becomes $(a!)(b!)=a!+(b+2)(b!)$, $a!(b!-1)=(b+2)(b!)$, since $(b!-1, b!)=1$, so $(b!-1) \mid (b+2)$, because $b$ is even, so $b=2, a!=8$ is impossible. In summary, the original equation has a unique solution $a=3, b=3, c=4$.