2025 AIME II Problem 10

Below is the professionally curated solution for Problem 10 of the 2025 AIME II, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AIME II solutions, or check the answer key.

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Concepts:arrangements with restrictionscombinationscasework

Difficulty rating: 2790

10.

Sixteen chairs are arranged in a row. Eight people each select a chair in which to sit so that no person sits next to two other people. Let NN be the number of subsets of 1616 chairs that could be selected. Find the remainder when NN is divided by 1000.1000.

Solution:

A person sits next to two others exactly when three consecutive chairs are all occupied, so we count 88-element subsets of the 1616 chairs with no three consecutive chairs chosen. The occupied chairs then form maximal blocks of size 11 or 2.2. If there are mm blocks, then 8m8 - m of them are pairs and 2m82m - 8 are singles, so 4m8,4 \le m \le 8, and the pair positions can be chosen in (m8m)\binom{m}{8-m} ways. The 88 empty chairs create 99 gaps (including the ends), and the mm blocks occupy mm distinct gaps: (9m)\binom{9}{m} ways.

Therefore N=m=48(m8m)(9m)=1126+10126+1584+736+19=2907.N = \sum_{m=4}^{8} \binom{m}{8-m}\binom{9}{m} = 1 \cdot 126 + 10 \cdot 126 + 15 \cdot 84 + 7 \cdot 36 + 1 \cdot 9 = 2907.

The remainder when N=2907N = 2907 is divided by 10001000 is 907.907.

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