If $(a+b)! = 4(b+c)! + 18(a+c)!$ then $(a+b)! > (a+c)!$, from which it follows that $b>c$. Similarly, $(a+b)! > (b+c)!$, so that $a>c$. Suppose for the moment that $b\geqslant a$, and let $x = \dfrac{(a+b)!}{(b+c)!}$, $y = \dfrac{(b+c)!}{(a+c)!}$. Then $x$ and $y$ are positive integers. After dividing through by $(a+c)!$, the factorial equation becomes $ \dfrac{(a+b)!}{(a+c)!} = \dfrac{4(b+c)!}{(a+c)!} + 18$, or $xy = 4y + 18$. Therefore $y(x-4) = 18$, and $y$ must be a factor of $18$. But not every factor of $18$ will lead to a solution of the equation, because $x$ and $y$ are defined in terms of factorials and so have to be products of consecutive integers. I found that the only values of $(x,y)$ that work are $(7,6)$ and $(22,1)$, corresponding to the solutions $(a,b,c) = (3,4,2)$ and $(11,11,10)$.
There seems to be no obvious reason why $b\geqslant a$, so we should also look at the possibility $a>b$. Then a similar calculation to the one above leads to an equation like $xy = 4y + 18$, but with the $4$ and $18$ interchanged. However, that did not lead to any new solutions of the factorial equation. So there are only two solutions, namely
$7! = 4\cdot6! + 18\cdot5!$ (when $(a,b,c) = (3,4,2)$)
and
$22! = 4\cdot21! + 18\cdot21!$ (when $(a,b,c) = (11,11,10)$).