Finding all integer solutions to $A\times B\times C = A! + B!+ C!$ with $0\leq A, B, C\leq 9$ without trying every combination? 
Let $A$, $B$, $C$ be integers such that  $0 \le A, B, C \le 9$. Find all the solutions for $$A\times B\times C = A! + B!+ C!$$

I have tried some values for $A$, $B$, and $C$ and found a solution: $A=4$, $B=3$, $C=3$.

Is there a way to solve this without using a computer to try every combination possible?


I don't think you can do it without some testing, but you don't have to try every possibility.  We can assume $$A\geq B\geq C$$
The right-hand side is greater than $A!$.  Dividing through by $A$ gives $BC>(A-1)!$, so certainly $81>(A-1)!$ and $A<6$.  
What if $A=5$?  Then we have, as above $$BC>24$$ and $C\leq B\leq A=5$, so $A=B=C=5$, but this isn't a solution.
So now we know $A\leq4$, and we proceed in the same way.  
