Words from distinct letters with a vowel requirement: From the letters {a, b, c, d, e, f}, how many 3-letter words (ordered arrangements, no repetition) can be formed that contain at least one vowel?

Introduction / Context:
We are forming ordered arrangements (permutations) of length 3 from 6 distinct letters, under the condition that each word contains at least one vowel. Using the complement (no vowels) simplifies the count substantially.

Given Data / Assumptions:

• Alphabet set: {a, b, c, d, e, f} with vowels a, e.
• Length = 3; no repetition; order matters.
• At least one vowel must appear in the arrangement.

Concept / Approach:

• Total 3-permutations from 6 letters = 6P3.
• Subtract arrangements with no vowels: choose only from {b, c, d, f} = 4P3.
• Use inclusion–exclusion via complement.

Step-by-Step Solution:

Verification / Alternative check:
Direct casework (1 vowel, 2 vowels) is longer but produces the same sum as the complement method; complement is cleanest here.

Why Other Options Are Wrong:

• 72 and 48 are partial counts from incomplete casework.
• “None of these” is false because 96 is achievable by complement counting.

Common Pitfalls:

• Treating the problem as combinations instead of permutations.
• Forgetting to exclude the no-vowel words entirely.

Final Answer:
96