Question 6 • Mathematics II
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.
AI-written and checked against the verified answer.