Equation involving factorials Find all integers a,b,c satisfying
$$a!b! = a! + b! + c!$$
Without loss of generality, we can write $b$ as $a-k$ (k is a positive integer), so we get
$$c! = a!(a-k)! - (a! + (a-k)!)$$
and so $c!$ must be divisible by $(a-k)!$, it follows that $c! = (a-k+r)!$ (where $r$ is a positive integer). I don't know what to do next. Not sure if what I have done is even on the right track, seems like all I have done is replaced $b,c$ by $k,r$

WOLOG let $a \ge b$.  We can't have $b=1$ as the equation becomes $0=1+c!$, nor $b=2$ as the equation becomes $a!=2+c!$
$b \ge 3, b! \ge 6,$ and $c \gt a$
Then if $b \lt a$ we can divide by $a!$ and get $b!=1+\frac {b!}{a!} + \frac {c!}{a!}$ which fails because $\frac {b!}{a!}$ is not an integer.
So $a=b$ and the equation becomes $(a!)^2=2(a!)+c!$ or $a!=2+\frac {c!}{a!}$
The left side is a multiple of $3$, so $c$ cannot be greater than $a+2$.
We can solve $a!=2+(a+1)$ for $a=3$, giving $3!3!=3!+3!+4!$, but $a$ cannot be greater as the factorial grows too fast.
There are no solutions to $a!=2+(a+1)(a+2)=a^2+3a+4$ as the left side is too small at $a=4$ and too big at $a=5$ 
The only solution is $3!3!=3!+3!+4!$
