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Explanation:

The value of ABC, three-digit number is should satisfy the condition 100A + 10B + C = A! + B! + C! ....(i)
The maximum value of three digit number is 999 and minimum is 100
We observe that 7! = 5040 > 999
So the 3-digits of the number must be
           6 as 6! = 720
and/or 5 as 5! = 120
and/or 4 as 4! = 24
and/or 3 as 3! = 6
and/or 2 as 2! = 4
and/or 1 as 1! = 1
If we consider 6 at the hundred’s place digit we see that condition (1) is not satisfied as 600 < 720 (6!) So we conclude that 6 cannot occupy any position in the number.
If we place ‘5’ at the hundred’s place then the number should lie between the range of 500 and 600. Considering the RHS of equation (1) by putting A + B = C = 5 we get the sum as 360 which is less than 500. Similarly, putting 4, 3, 2 at the hundred’s place does not satisfies the given condition (1).
Only 1 can be placed at hundred place and 5 should be one of the digit at other two position in order to make it a three digit number.

Thus, only combination we satisfies the given condition (1) is (1, 4, 5) i.e. 145 = 1! + 4! + 5! = 145.