Three, (50 points) Find all triples of positive integers $(a, b, c)$ such that $a, b, c$ satisfy
$$
(a!)(b!)=a!+b!-c!.
$$
There are $\frac{a!}{b!}$ and $\frac{c!}{b!}$, and only one of them is odd.
Below, we discuss the scenarios.
(1) Assume $\frac{a!}{b!}$ is odd and $\frac{c!}{b!}$ is even, then either $a=b$, or $\frac{a!}{b!}=b+1$, and $b+1$ is odd, $a=b+1$.
(i) If $a=b$, then $a!=2-\frac{c!}{a!}b$, hence $(b+1) \left\lvert\, \frac{c!}{b!}\right.$. Thus, $(b+1) \mid (b+2)$, which is a contradiction.
(2) Assume $\frac{a!}{b!}$ is even and $\frac{c!}{b!}$ is odd, then $c=b$ or $b+1$ (where $b$ is even).
(i) If $c=b$, then the original equation becomes
$(a!)(b!)=a!$.
Thus, $b!=1 \Rightarrow b=1, c=1, a \geqslant 3$.
(ii) If $c=b+1$, then the original equation becomes $(a!)(b!)=a!-b(b!)$.
Notice that, $(a!)(b!) \geqslant a!, a!-b(b!)<a!$. This scenario is impossible.
In summary, $(a, b, c)=(m, 1,1),(1, m, 1)\left(m \in \mathbf{Z}_{+}\right)$.
Three, without loss of generality, assume $a \geqslant b$.
When $a=1$, $b=1$, then $c!=1 \Rightarrow c=1$.
When $a=2$, $b=1$ or 2, only when $b=1, c=1$ does the original equation hold.
When $a \geqslant 3$, $a!$ is even, the original equation becomes $a!=\frac{a!}{b!}+1-\frac{c!}{b!}$.
Since all three terms in the above equation are integers, then $c \geqslant b$.