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ABC = A! + B! + C!
Cannot have 6, 7, 8, 9 as digits
At least one digit to be equal to 5
Exactly one digit is equal to 5
I don't think there is a solution for this using the "factorial function"!. However, there is a solution using the "Gamma function". It gives the following solution:
A*B*C =A! + B! + C!, solve for A, B, C
where: A ≈ 4.39056.... and B = -53/10 ≈ -5.30000 and C = -17/10 ≈ -1.70000
Note: This is courtesy of: Mathematica 11 Home Edition.