Question 2 • TN-Algebra II B SC 26-27 (For Credit)
Since zero is not allowed, the code’s digits come from 1 through 9. There are 9 × 8 × 7 = 504 possible codes when no digit is repeated. The available even digits are 2, 4, 6, and 8, so the number of three-digit codes using distinct even digits is 4 × 3 × 2 = 24. Thus the probability is 24 out of 504, choice A.
AI-written and checked against the verified answer.