2025 AIME II 第 10 题

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10.

十六把椅子排成一排。八个人各选择一把椅子坐下,并且没有人同时紧挨着另外两个人。令 NN 为可能被选中的 1616 把椅子的子集数量。求 NN 除以 10001000 的余数。

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.

答案:907
知识点:有限制的排列组合分类讨论
难度评级:2790
解答:

一个人同时紧挨着另外两个人,恰好等价于三把连续椅子都被占用,所以我们要数没有三把连续椅子被选中的 1616 把椅子的 88 元子集。被占用的椅子于是形成大小为 1122 的极大区块。 若有 mm 个区块,则其中 8m8 - m 个是二连座,2m82m - 8 个是单座,所以 4m84 \le m \le 8,二连座的位置可用 (m8m)\binom{m}{8-m} 种方式选择。88 把空椅子形成 99 个空隙(包括两端),mm 个区块占据 mm 个不同空隙:有 (9m)\binom{9}{m} 种方式。

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

N=2907N = 2907 除以 10001000 的余数为 907907

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. \begin{gathered} N = \sum_{m=4}^{8} \binom{m}{8-m}\binom{9}{m} \\ = 1 \cdot 126 + 10 \cdot 126 + 15 \cdot 84 \\ \quad {}+ 7 \cdot 36 + 1 \cdot 9 = 2907. \end{gathered}

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

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