2025 AMC 12A Problem 23

Below is the professionally curated solution for Problem 23 of the 2025 AMC 12A, 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.

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Concepts:arrangements with restrictionsbinomial theorem

Difficulty rating: 2340

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

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 2n12^{n-1} valid arrangements.

Summing over all nonempty digit subsets, n=19(9n)2n1=12n=19(9n)2n=3912=196822=9841.\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.

Thus, the correct answer is C.

Problem 23 in Other Years