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How many three digit numbers with distinct digits can be formed such that product of the digits is the cube of a positive integer? a)21 b)24 c)36 d)30

Precalculus
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Let one digit is fixed to be 1. Then product of two distinct digits must be a perfect cube i.e. (2,4) such that 1*2*4 = 8 = 23 total 3! = 6 combinations (3,9) such that 1*3*9 = 27 = 33 total 3! = 6 combinations One digit is fixed to be 2. (4,8) such that 2*4*8 = 64 = 43 total 3! = 6 combinations One digit is fixed to be 3. (8,9) such that 3*8*9 = 216 = 63 total 3! = 6 combinations One digit is fixed to be 4 (6,9) such that 4*6*9 = 216 = 63 total 3! = 6 combinations
thanks @zpupster
@zpupster I got you except 23 total, 33 total, 43 total.... what do they mean, please explain me. btw , your explanation is perfect

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