A briefcase has a three-digit lock code that does not include zero as a digit. What is the probability that the lock code consists of all even digits if the same digit is not used more than once in the lock code?

There are 9 possible nonzero digits, and the three positions must use different digits, so there are 9 × 8 × 7 = 504 possible codes. The even nonzero digits are 2, 4, 6, and 8; choosing and ordering three distinct ones gives 4 × 3 × 2 = 24 all-even codes. Therefore the probability is 24 out of 504, choice B.

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