Question

亚瑟王的二十五名骑士围坐在他们惯用的圆桌旁。从中随机选出三名骑士去消灭一条惹麻烦的龙，每组三人的组合被选中的可能性相同。设 $P$ 为所选三人中至少有两人原本相邻而坐的概率。若将 $P$ 写成最简分数，求其分子与分母之和。

Twenty-five of King Arthur’s knights are seated at their customary round table. Three of them are chosen, with all choices of three equally likely, and are sent off to slay a troublesome dragon. Let $P$ be the probability that at least two of the three had been sitting next to each other. If $P$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

答案

答案：57

知识点：组合补集计数对立事件概率

难度评级：2410

小提示：

计算任意两名所选骑士都不相邻的补集选法数

Count the complementary selections in which no two chosen knights are adjacent

大提示：

分成包含某个固定骑士的选法和不包含该骑士的选法

Split into selections containing a fixed knight and selections not containing that knight

解答：

共有 $\binom{25}{3}=2300$ 种选法。固定一个座位。若不选该座位，则在其余 $24$ 个座位中选出三个互不相邻的座位，等价于从一列 $24$ 个位置中选三个不连续的位置，共有 $\binom{22}{3}$ 种。若选定该固定座位，则它的两个邻座不能选，而另外两个座位必须在剩余的一列 $22$ 个位置中互不连续，共有 $\binom{21}{2}$ 种。

因此，所选座位中没有相邻座位的选法数为 $\begin{aligned} \binom{22}{3}+\binom{21}{2} &=1540+210\\ &=1750 \end{aligned}\text{。}$ 所以 $P=1-\frac{1750}{2300}=\frac{11}{46}\text{，}$ 所求和为 $11+46=57$ 。

There are $\binom{25}{3}=2300$ selections. Fix one seat. If it is not selected, choosing three nonadjacent seats among the remaining $24$ seats is equivalent to choosing three nonconsecutive positions from a row of $24,$ giving $\binom{22}{3}.$ If the fixed seat is selected, its two neighbors are forbidden, and the other two selected seats must be nonconsecutive among the remaining row of $22,$ giving $\binom{21}{2}.$

Thus the number with no adjacent selected seats is $\begin{aligned} \binom{22}{3}+\binom{21}{2} &=1540+210\\ &=1750. \end{aligned}$ Therefore $P=1-\frac{1750}{2300}=\frac{11}{46},$ and the requested sum is $11+46=57.$

1983 AIME 第 7 题

答案：57

难度评级：2410

小提示：

大提示：

解答：

其他年份的第 7 题