AMC10 2021 B

AMC10 2021 B · Q22

AMC10 2021 B · Q22. It mainly tests Permutations, Combinations.

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 $m + n ?$

Ang、Ben和Jasmin各有5块积木，颜色分别为红、蓝、黄、白、绿；有5个空盒子。三人独立随机地将各自的一块积木放入每个盒子中。至少有一个盒子收到3块同色积木的概率为$\frac{m}{n}$，其中$m$和$n$互质。求$m+n$？

(A) 47 47

(B) 94 94

(D) 471 471

(E) 542 542

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let our denominator be $(5!)^3$, so we consider all possible distributions. We use PIE (Principle of Inclusion and Exclusion) to count the successful ones. When we have at $1$ box with all $3$ blocks the same color in that box, there are $_{5} C _{1} \cdot _{5} P _{1} \cdot (4!)^3$ ways for the distributions to occur ($_{5} C _{1}$ for selecting one of the five boxes for a uniform color, $_{5} P _{1}$ for choosing the color for that box, $4!$ for each of the three people to place their remaining items). However, we overcounted those distributions where two boxes had uniform color, and there are $_{5} C _{2} \cdot _{5} P _{2} \cdot (3!)^3$ ways for the distributions to occur ($_{5} C _{2}$ for selecting two of the five boxes for a uniform color, $_{5} P _{2}$ for choosing the color for those boxes, $3!$ for each of the three people to place their remaining items). Again, we need to re-add back in the distributions with three boxes of uniform color... and so on so forth. Our success by PIE is \[_{5} C _{1} \cdot _{5} P _{1} \cdot (4!)^3 - _{5} C _{2} \cdot _{5} P _{2} \cdot (3!)^3 + _{5} C _{3} \cdot _{5} P _{3} \cdot (2!)^3 - _{5} C _{4} \cdot _{5} P _{4} \cdot (1!)^3 + _{5} C _{5} \cdot _{5} P _{5} \cdot (0!)^3 = 120 \cdot 2556.\] \[\frac{120 \cdot 2556}{120^3}=\frac{71}{400},\] yielding an answer of $\boxed{\textbf{(D) }471}$. As In Solution 1, the probability is \[\frac{\binom{5}{1}\cdot 5\cdot (4!)^3 - \binom{5}{2}\cdot 5\cdot 4\cdot (3!)^3 + \binom{5}{3}\cdot 5\cdot 4\cdot 3\cdot (2!)^3 - \binom{5}{4}\cdot 5\cdot 4\cdot 3\cdot 2\cdot (1!)^3 + \binom{5}{5}\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{(5!)^3}\] \[= \frac{5\cdot 5\cdot (4!)^3 - 10\cdot 5\cdot 4\cdot (3!)^3 + 10\cdot 5\cdot 4\cdot 3\cdot (2!)^3 - 5\cdot 5! + 5!}{(5!)^3}.\] Dividing by $5!$, we get \[\frac{5\cdot (4!)^2 - 10\cdot (3!)^2 + 10\cdot (2!)^2 - 5 + 1}{(5!)^2}.\] Dividing by $4$, we get \[\frac{5\cdot 6\cdot 24 - 10\cdot 9 + 10 - 1}{30\cdot 120}.\] Dividing by $9$, we get \[\frac{5\cdot 2\cdot 8 - 10 + 1}{10\cdot 40} = \frac{71}{400} \implies \boxed{\textbf{(D) }471}.\]

设分母为$(5!)^3$，考虑所有可能的分配。我们使用PIE（容斥原理）来计算成功的分配数。当有$1$个盒子有3块同色积木时，有$\binom{5}{1} \cdot {}_5P_1 \cdot (4!)^3$种方式（选择5个盒子中的一个用于统一颜色${}_5C_1$，选择该盒子的颜色${}_5P_1$，三人放置剩余物品各$4!$）。但是，我们多计了两个盒子有统一颜色的分配，有$\binom{5}{2} \cdot {}_5P_2 \cdot (3!)^3$种方式（选择两个盒子${}_5C_2$，选择颜色${}_5P_2$，三人放置剩余各$3!$）。同样，需要加回三个盒子统一颜色的分配……以此类推。通过PIE的成功数为 \[\binom{5}{1} \cdot {}_5P_1 \cdot (4!)^3 - \binom{5}{2} \cdot {}_5P_2 \cdot (3!)^3 + \binom{5}{3} \cdot {}_5P_3 \cdot (2!)^3 - \binom{5}{4} \cdot {}_5P_4 \cdot (1!)^3 + \binom{5}{5} \cdot {}_5P_5 \cdot (0!)^3 = 120 \cdot 2556.\] \[\frac{120 \cdot 2556}{120^3}=\frac{71}{400},\] 得答案$\boxed{\textbf{(D) }471}$。如解1，概率为 \[\frac{\binom{5}{1}\cdot 5\cdot (4!)^3 - \binom{5}{2}\cdot 5\cdot 4\cdot (3!)^3 + \binom{5}{3}\cdot 5\cdot 4\cdot 3\cdot (2!)^3 - \binom{5}{4}\cdot 5\cdot 4\cdot 3\cdot 2\cdot (1!)^3 + \binom{5}{5}\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{(5!)^3}\] \[= \frac{5\cdot 5\cdot (4!)^3 - 10\cdot 5\cdot 4\cdot (3!)^3 + 10\cdot 5\cdot 4\cdot 3\cdot (2!)^3 - 5\cdot 5! + 5!}{(5!)^3}.\] 除以$5!$，得 \[\frac{5\cdot (4!)^2 - 10\cdot (3!)^2 + 10\cdot (2!)^2 - 5 + 1}{(5!)^2}.\] 除以$4$，得 \[\frac{5\cdot 6\cdot 24 - 10\cdot 9 + 10 - 1}{30\cdot 120}.\] 除以$9$，得 \[\frac{5\cdot 2\cdot 8 - 10 + 1}{10\cdot 40} = \frac{71}{400} \implies \boxed{\textbf{(D) }471}.\]

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