AMC 10 · 2025 · #24

Grade 8 counting
Archetype Branching State Enumeration └ Sub-archetype Constrained Sequence or Path Count
combinations-basiccombinatorial-identitysystematic-enumeration convert-to-algebra ↑ Prerequisites: combinations-basic
📏 Medium solution 💡 3 insights
Problem
A positive integer is called fair when all of its digits are different, none of them is 0, and no digit has two neighbors that are both larger than it. Count how many fair positive integers exist.

Pick an answer.

(A)
511
(B)
2584
(C)
9841
(D)
17711
(E)
19682

AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

This is a 'how many numbers' count over a messy adjacency rule, so Tool #2 (Make a Systematic List) drives it: organize every fair number by which set of digits it uses. Tool #16 (Change Focus) supplies the key reframe — 'no digit between two larger ones' is the same as 'the digits climb to one peak then fall,' turning a checking rule into a shape. Tool #7 (Identify Subproblems) splits the job into 'count the arrangements for one fixed digit set' then 'add over all digit sets.' Tool #5 (Look for a Pattern) recognizes the resulting sum as a binomial expansion so it collapses to a clean 3⁹.

1STEP 1

Turn the rule into a shape

Only interior digits can be a valley, so a fair number must climb to one peak and then fall — a mountain of distinct nonzero digits.

fair ⇔ d₁ ＜ d₂ ＜ … ＜ d_peak ＞ … ＞ d_n-1 ＞ d_n
2STEP 2

Count mountains for one digit set

Fix k digits: the largest is forced to the peak and each of the other k-1 digits only picks a slope, so one set gives 2^(k-1) mountains.

2 × 2 × … × 2 = 2^ k-1 mountains per k-digit set
3STEP 3

Add over every digit set

There are C(9,k) ways to choose which k digits to use, so add C(9,k)·2^(k-1) over k from 1 to 9 to cover every length.

total=Σ_k=1⁹C(9, k) 2^ k-1
4STEP 4

Collapse the sum and finish

Halving and the binomial theorem fold the sum into half of 3⁹−1, giving 9841 — choice (C), while the unhalved 19682 is trap (E).

Σ_k=1⁹C(9, k)2^k-1=1/2(3⁹-1)=19682/2=9841 → (C)
Answer
9841
The count must land between two easy anchors. There are 9 one-digit fair numbers, and using all nine digits gives 2⁸=256 mountains, with many sizes in between, so a total in the thousands is sensible — 9841 fits while tiny (A) 511 does not. The two large traps confirm the method: 19682=3⁹-1 (choice E) drops the necessary 1/2, and doubling to 17711 (choice D) is a Fibonacci lookalike unrelated to this count. Only 9841=(3⁹-1)/2 respects that the peak digit is fixed while the other k-1 digits each choose a side.
💡Key takeaway

If distinct digits can never dip in the middle, the number must climb to one peak and slide back down — so just fix the biggest digit on top and let every other digit pick a side.

  • Turn the rule into a shape
  • Count mountains for one digit set
  • Add over every digit set
  • Collapse the sum and finish

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