Question

一位园丁把三棵枫树、四棵橡树和五棵桦树种成一排。他按随机顺序种植，每种排列出现的可能性相同。若任意两棵桦树都不相邻的概率以最简分数表示为 $\frac{m}{n}$ ，求 $m+n$ 。

A gardener plants three maple trees, four oak trees, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let $\frac{m}{n}$ in lowest terms be the probability that no two birch trees are next to one another. Find $m+n.$

答案

答案：106

知识点：有限制的排列组合基本概率

难度评级：2160

小提示：

先选择桦树占据的五个位置

First choose the five positions occupied by birch trees

大提示：

先排好七棵非桦树，再利用它们周围的八个空隙

Place the seven non-birch trees first and use the eight gaps around them

解答：

五棵桦树的位置构成一个 $5$ 元子集，它是从 $12$ 个位置中等可能选出的，因此共有 $\binom{12}{5}$ 种可能。排好七棵非桦树后，包括两端的空隙在内共有八个空隙。从中选取五个不同的空隙，可得 $\binom85$ 种没有相邻桦树的排列。因此 $\frac{m}{n}=\frac{\binom85}{\binom{12}{5}} =\frac{56}{792}=\frac7{99}\text{，}$ 所以 $m+n=106$ 。

The five birch positions form a uniformly chosen $5$ -element subset of the $12$ positions, so there are $\binom{12}{5}$ possibilities. After the seven non-birch trees are placed, there are eight gaps, including the two end gaps. Choosing five distinct gaps gives $\binom85$ arrangements with no adjacent birches. Therefore $\frac{m}{n}=\frac{\binom85}{\binom{12}{5}} =\frac{56}{792}=\frac7{99},$ and $m+n=106.$

1984 AIME 第 11 题

答案：106

难度评级：2160

小提示：

大提示：

解答：

其他年份的第 11 题