Question 313391
Firstly, we have 120 = 2^3*3*5
since each digit is smaller than 10, therefore, one digit must be 5. We have 2 digits left with product of 2^3*3
there are only two case for one digit that has the factor of 3: 3, 3*2 (3*2^2 is larger than 9)
So there are totally two 3-digit sets that has the product of 120: (5,3,8) and (5,6,4)
With each set, we have 3! different digits that satisfied the product of the three digits is 120.
Thus, there are totally 2*3! = 12 such digits.