2025 AMC 12A Problem 23

Attempt Problem 23 of the 2025 AMC 12A 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 2025 AMC 12A solutions, or check the answer key.

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

23.

Call a positive integer fair if no digit is used more than once, it has no 00s, and no digit is adjacent to two greater digits. For example, 196,196, 23,23, and 1246312463 are fair, but 1546,1546, 320,320, and 3432134321 are not fair. How many fair positive integers are there?
511511
25842584
98419841
1771117711
1968219682
Answer: C
Concepts:arrangements with restrictionsbinomial theorem
Small hint:
A fair number has no interior digit smaller than both of its neighbors
Big hint:
Build a valid arrangement of a chosen digit set by placing digits largest-to-smallest, each at either end: 2n−12^{n-1} ways for nn digits
Solution:
The digits are distinct and drawn from {1,…,9},\{1, \ldots, 9\}, and “no digit adjacent to two greater digits” means no interior digit is smaller than both neighbors.
For a fixed set of nn digits, build the arrangement by inserting digits from largest to smallest; each new (smaller) digit must go to one of the two ends, giving 2n−12^{n-1} valid arrangements.
Summing over all nonempty digit subsets, ∑n=19(9n)2n−1=12∑n=19(9n)2n=39−12=196822=9841. \begin{gathered} \sum_{n=1}^{9}\binom{9}{n}2^{n-1} \\ = \frac{1}{2}\sum_{n=1}^{9}\binom{9}{n}2^{n} \\ = \frac{3^9 - 1}{2} = \frac{19682}{2} = 9841. \end{gathered}
Thus, the correct answer is C.
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