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Since $ABC$ is a three digit number, we cannot have $7$ as one of its digits as $7! = 5040$ is a four digit number .Similarly , we cannot have $8,9.$

We cannot even have $6$ as $6! = 720$ and when you add any other factorial, it will have $7 , 8$ or $9$ as one of its third digit i.e $A= 7,8$ or $9$ . We know that we cannot have $7 , 8, 9$ as digits.

Now, we have to take one digit as $5$ compulsory as if we exclude $5$ we can a maximum value of $ABC$ when $A = B = C = 4 .$

$4! + 4! + 4! = 72$ which is not even a three digit number.

When we take one of the digits as $5$ we can get maximum value when others are $4$ i.e, $4! +4! + 5!= 168 \neq 445.$

From here we can conclude that one of the digits is $1$

Therefore, $A! + B! + C! = 1! + B! + 5! = 1B5$

Use trial and error for $B$ in between $1$ to $4$ . The condition is satisfied for $B = 4$

$\boxed{A! + B! + C! = 1! + 4! + 5! = {\color{#20A900}{145} }= ABC}$

If the largest digit of $ABC$ was greater than or equal to 6, its factorial would be greater than or equal to 720, but then the largest digit would have to be at least 7, which contradicts the assumption and therefore doesn't work.

Let's try the largest digit to be 5. $5!=120$ , so the first digit must be 1. $1!+5!=121$ . Now we can check and find that 4 works, since $1!+4!+5! = \boxed{145}$ .