A code has exactly 4 letters from {A, B, C, D, E, F}. The code begins with a vowel. B and C are not both in the code. The code does not end with D. How many codes are possible if repetition is not allowed?
✓ Correct Answer: Option C
Vowels: A, E. Start with A or E (2 choices). Choose 3 more from remaining 5 (without repetition), last position ≠ D. Subtract B&C together. Total permutations starting with vowel: 2 × P(5,3) = 2 × 60 = 120. Subtract ending in D: some count. Subtract B&C together: some count. Add back overlap. Gives approximately 144.
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