Question

Ang、Ben 和 Jasmin 每人都有 $5$ 块积木，颜色分别为红、蓝、黄、白、绿；另有 $5$ 个空盒子。三个人各自随机且彼此独立地把自己的每种颜色积木各放入一个盒子中。至少有一个盒子收到 $3$ 块同色积木的概率为 $\frac{m}{n}$ ，其中 $m$ 与 $n$ 是互质正整数。 $m + n$ 等于多少？

Ang, Ben, and Jasmin each have $5$ blocks, colored red, blue, yellow, white, and green; and there are $5$ empty boxes. Each of the people randomly and independently of the other two people places one of their blocks into each box. The probability that at least one box receives $3$ blocks all of the same color is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. What is $m + n ?$

$47$

$94$

$227$

$471$

$542$

答案：D

知识点：容斥原理排列基本概率

难度评级：2150

视频讲解：

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文字解答：

固定 Ang 的摆放方式，并用 Ang 放入的颜色标记每个盒子。Ben 和 Jasmin 各自选择五种颜色的一个排列，因此共有 $(5!)^2$ 个等可能的摆放对。

若指定 $k$ 个盒子都收到三块同色积木，则 Ben 和 Jasmin 在这 $k$ 个盒子中都必须与 Ang 的颜色相同。这可以用 $((5-k)!)^2$ 种方式完成。由容斥原理，成功的摆放对数为

$\begin{aligned} &\binom51(4!)^2-\binom52(3!)^2 \\ &\quad {}+\binom53(2!)^2-\binom54(1!)^2 \\ &\quad {}+\binom55(0!)^2. \end{aligned}$

上式等于

$2880-360+40-5+1=2556.$

这等于

$\frac{2556}{(5!)^2}=\frac{2556}{14400}=\frac{71}{400}.$

因此 $m+n=71+400=471$ 。

所以答案是 D。

Fix Ang's placement and label each box by the color Ang put in it. Ben and Jasmin each choose a permutation of the five colors, so there are $(5!)^2$ equally likely pairs of placements.

For a specified set of $k$ boxes to receive three blocks of the same color, both Ben and Jasmin must match Ang in those $k$ boxes. This can happen in $((5-k)!)^2$ ways. By inclusion-exclusion, the number of successful placement pairs is

$\begin{aligned} &\binom51(4!)^2-\binom52(3!)^2 \\ &\quad {}+\binom53(2!)^2-\binom54(1!)^2 \\ &\quad {}+\binom55(0!)^2. \end{aligned}$

This equals

$2880-360+40-5+1=2556.$

Therefore the probability is

$\frac{2556}{(5!)^2}=\frac{2556}{14400}=\frac{71}{400}.$

Thus $m+n=71+400=471.$

Thus, the answer is D .

2021 AMC 10B Spring 第 22 题

答案：D

难度评级：2150

视频讲解：

文字解答：

其他年份的第 22 题