Three-letter passwords (no repetition) with at least one symmetric letter: Assume symmetric letters about a vertical axis are A, H, I, M, O, T, U, V, W, X, Y (11 letters). How many 3-letter passwords (A–Z, no repetition) contain at least one symmetric letter?

Introduction / Context:We work with uppercase English letters. A standard convention for “symmetric letters” (mirror symmetry about a vertical axis) is {A, H, I, M, O, T, U, V, W, X, Y}. We count 3-letter permutations (no repetition) containing at least one such letter using the complement method.

Given Data / Assumptions:

• Total letters = 26; symmetric set size = 11; nonsymmetric = 15.
• Length = 3; no letter repeats; order matters.
• At least one symmetric letter required.

Concept / Approach:

• Total permutations = 26P3.
• No-symmetric permutations = 15P3.
• Desired = 26P3 − 15P3.

Step-by-Step Solution:

Verification / Alternative check:Direct case splits by number of symmetric letters (1, 2, 3) lead to the same total; complement is faster and less error-prone.

Why Other Options Are Wrong:

• 2730 is the complement (no symmetric letters).
• 1560000 assumes 26^3 with repetition, not 26P3.
• 990 is unrelated to this setup.

Common Pitfalls:

• Using combinations instead of permutations.
• Assuming repetition allowed when it is prohibited.

Final Answer:12870