1989 AMC 12 Problem 9

Attempt Problem 9 of the 1989 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1989 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

9.

Mr. and Mrs. Zeta want to name their baby Zeta so that its monogram (first, middle, and last initials) will be in alphabetical order with no letters repeated. How many such monograms are possible?
276276
300300
552552
600600
1560015600
Answer: B
Concepts:combinationsarrangements with restrictions
Small hint:
The last initial is already fixed as ZZ
Big hint:
Choosing the other two distinct letters determines their alphabetical order
Solution:
The first two initials must be two distinct letters chosen from AA through Y.Y. Once chosen, their order is forced. Therefore the number of monograms is (252)=300. \binom{25}{2}=300.
Thus the correct answer is B.
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