let $a \leq b$ $a =1$ - there are not solutions $a=2: b!=2+c!$ there are not solutions If $a\geq 3$ then $c!=a!b!-a!-b! \geq 4*b! \to c>b$ and so $b! | a!*b!-b!-c!=a! \to b=a$ $a!^2=2*a!+c!$ $a!=2+\frac{c!}{a!} $ If $c \geq a+4$ then $4| \frac{c!}{a!} \to a! \equiv 2 \pmod{4} \to a=3$ but there are not solutions with $c \geq a+4$ $c=a+1: a!=2+a+1: a=3,b=3,c=4$ $c=a+2: a!=2+(a+1)(a+2) \to a|4 \to a=4$ but there are not solutions $c=a+3: a!=2+(a+1)(a+2)(a+3) \to a|8 \to a=4,8$ but there are not solutions So $(a,b,c)=(3,3,4)$

https://www.youtube.com/watch?v=9dyK_op-Ocw British MO

Case 1: $a=1$ The equation transforms into $b!=1+b!+c!\Longrightarrow c!+1=0$ which is clearly a contradiction, so there are no solutions for $a=1$. Case 2: $a=2$ The equation becomes $2b!=2+b!+c!\Longrightarrow b!=2+c!\Longleftrightarrow c\ge2$ furthermore if we take$\pmod 4$, we notice that $c!\equiv 2\pmod 4$ or in other words $c=2$ or $c=3$ Subcase 1: $c=3$ $b!=2+3!=8$ which is a contradiction. Subcase 2: $c=2$ $b!=2+2!=4$ which is a contradiction. Thus there are no solution when $a=2$ Case 3: $a\ge3$ and $WLOG$ assume $b\ge a$ Notice that we can rewrite the equation like $c!=a!b!-a!-b!\ge 4b!$ from the fact that $a\ge3$, thus $c>b$. Furthermore notice that since $c>b$, $b!|a!b!-b!-c!=a!$ thus $a\ge b$, which forces $a=b$ from out assumption. Therefore the equation becomes $2a!+c!=a!^2\Longrightarrow 2+\frac{c!}{a!}=a!$, furthermore notice that if $c\ge a+4$, $4|\frac{c!}{a!}\Longrightarrow a!\equiv 2\pmod 4$ or $a\le 3$, thus $c\le a+3$. Subcase 1: $c=a+3$ $a!=a^3+6a^2+11a+8\Longleftrightarrow (a-1)!=a^2+6a+11=\frac{8}{a}$ thus $a\le8$. A manual check reveals no solutions. Subcase 2: $c=a+2$ $a!=a^2+3a+4\Longrightarrow (a-1)!=a+3+\frac{4}{a}\Longleftrightarrow a=4$, after checking we obtain that this in not a solution. Subcase 3: $c=a+1$ $a!=3+a\Longrightarrow (a-1)!=1+\frac{3}{a}\Longleftrightarrow a=3$, after a manual check we get that $a=3, b=3\text{ and } c=4$ is a solution $\blacksquare$. So, to sum up $\boxed{(a,b,c)=(3,3,4)}$ are the only solutions.