Question

Kathy 有 $5$ 张红卡和 $5$ 张绿卡。她洗混这 $10$ 张卡，并随机按顺序排出其中 $5$ 张。当且仅当所有排出的红卡相邻且所有排出的绿卡相邻时，她会满意。例如，卡片顺序 RRGGG、GGGGR 或 RRRRR 会让 Kathy 满意，但 RRRGR 不会。Kathy 满意的概率为 $\frac{m}{n}$ ，其中 $m$ 与 $n$ 是互质正整数。求 $m + n$ 。

Kathy has $5$ red cards and $5$ green cards. She shuffles the $10$ cards and lays out $5$ of the cards in a row in a random order. She will be happy if and only if all the red cards laid out are adjacent and all the green cards laid out are adjacent. For example, card orders RRGGG, GGGGR, or RRRRR will make Kathy happy, but RRRGR will not. The probability that Kathy will be happy is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：157

知识点：基本概率有限制的排列分类讨论

难度评级：2400

解答：

从 $10$ 张不同卡中排出 $5$ 张，共有 $10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 = 30240$ 个等可能的有序排列。Kathy 满意恰好意味着颜色模式由一个红色块和一个绿色块组成：模式包括 RRRRR、GGGGG，以及 $r = 1, 2, 3, 4$ 时的八种混合模式 $\text{R}^r\text{G}^{5-r}$ 和 $\text{G}^{5-r}\text{R}^r$ 。

一个使用 $r$ 个红色位置和 $5 - r$ 个绿色位置的模式，可以用 $\frac{5!}{(5-r)!} \cdot \frac{5!}{r!}$ 种方式填入具体卡片。对 $r = 5, 4, 3, 2, 1, 0$ ，这些数分别为 $120$ 、 $600$ 、 $1200$ 、 $1200$ 、 $600$ 、 $120$ 。满意的排列数为 $\begin{aligned} &120 + 120 \\ &{}+ 2\,(600 + 1200 + 1200 + 600) \\ &{}= 7440. \end{aligned}$

概率为 $\frac{7440}{30240} = \frac{31}{126}$ ，所以 $m + n = 31 + 126 = 157$ 。

There are $10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 = 30240$ equally likely ordered layouts of $5$ of the $10$ distinct cards. Kathy is happy exactly when the color pattern consists of one block of reds and one block of greens: the patterns are RRRRR, GGGGG, and the eight mixed patterns $\text{R}^r\text{G}^{5-r}$ and $\text{G}^{5-r}\text{R}^r$ for $r = 1, 2, 3, 4.$

A pattern using $r$ red and $5 - r$ green positions can be filled in $\frac{5!}{(5-r)!} \cdot \frac{5!}{r!}$ ways (ordered choices of which red cards and which green cards appear). For $r = 5, 4, 3, 2, 1, 0$ these counts are $120,$ $600,$ $1200,$ $1200,$ $600,$ $120.$ The happy layouts number $\begin{aligned} &120 + 120 \\ &{}+ 2\,(600 + 1200 + 1200 + 600) \\ &{}= 7440. \end{aligned}$

The probability is $\frac{7440}{30240} = \frac{31}{126},$ so $m + n = 31 + 126 = 157.$

2018 AIME I 第 3 题

答案：157

难度评级：2400

解答：

其他年份的第 3 题