Thats something interesting! (part 2)

Without loss of generality let us assume that $\large a \geq b$ and divide through by $\large b!$ :

$\large a! = \dfrac{a!}{b!} + 1 +\dfrac{2^c}{b!}$ , giving integers throughout.

As each term on the RHS is integer, $\large RHS \geq 3 \Rightarrow a!\geq 3 \Rightarrow a\geq 3$

For $\large b > 2$ , $\large b!$ will contain a factor of $\large 3$ , and so cannot divide $\large 2^c$ .

Thus $\large b=1 ,or, b=2$

If $\large b=1\Rightarrow a! = a! + 1 + 2^c$ , which leads to $\large 2^c + 1 = 0:$ no solution.

If $\large b=2,\Rightarrow a! = \dfrac{a!}{2} + 1 + 2^{c - 1}$ , which leads to $\large \dfrac{a!}{2} = 1 + 2^{c -1}$

If $\large a > 3 \Rightarrow \dfrac{a!}{2}$ is even, so $\large 2^{c - 1} = 1$ . But then we get $\large \dfrac{a!}{2} = 2$ no solution.

If $\large a = 3, 3 = 1 + 2^{c - 1} \Rightarrow c = 2$

Hence we obtain the only solution: $\large 3!2! = 3! + 2! + 2^2$

Thus, $\large a + b + c = 3 + 2 + 2 =\huge \boxed{7}$ .

Assume without loss of generality that $\large a \geq b$ . Now dividing by $\large b!$ gives $a! = \frac{a!}{b!} + 1 + \frac{2^c}{b!}$ This means that $\large b = 1 \ or \ 2$ . For $\large b =1$ we get $\large 1+2^c = 0$ so $\large b = 2$ $\Rightarrow$ $\large a! = \frac{a!}{2} + 1 + 2^{c-1} \Rightarrow\ 2^{c-1} + 1= \frac{a!}{2}$ . This means that $\large \frac{a!}{2}$ must be odd so the exponent of 2 in the factorial is 1. Hence, (as $2^{c-1}$ > 0 ) $\large a = 3$ and $\large c = 2$

$\therefore$ $\large a+b+c = 3+2+2 = \boxed{7}$

Done through hit and trial method...

just guess

$2!\times 3!=2!+3!\times { 2 }^{ 2 }$

Hence $2+3+2=7$