A password uses exactly 3 of {A, B, C, D, E}. The password must contain A or B (or both). C and D cannot both be in the password. How many valid passwords are there (order matters)?

✓ Correct Answer: Option D

Total 3-letter permutations from 5: P(5,3)=60. Without A and B: choose 3 from {C,D,E} = P(3,3)=6. So with A or B: 60-6=54. But wait, must also subtract C&D together cases. With C&D: choose 1 more from {A,B,E}: 3 choices, arrange 3: 3×3!=18 arrangements. Without A or B from those: just {C,D,E}=6, but already excluded. Hmm, 54 minus (C&D cases that have A or B): 18-6=12 to subtract? Let me recalculate... Valid = (has A or B) AND (not both C&D). Total with A or B = 54. Among those, with both C&D: pick from {A,B} for 3rd spot: 2 × 3! = 12. Valid = 54-12 = 42. answer is C.