AMC10 2025 B

AMC10 2025 B · Q14

AMC10 2025 B · Q14. It mainly tests Basic counting (rules of product/sum), Probability (basic).

Nine athletes, no two of whom are the same height, try out for the basketball team. One at a time, they draw a wristband at random, without replacement, from a bag containing 3 blue bands, 3 red bands, and 3 green bands. They are divided into a blue group, a red group, and a green group. The tallest member of each group is named the group captain. What is the probability that the group captains are the three tallest athletes?

九名身高均不同的运动员试训篮球队。他们依次从袋中随机抽取腕带，不放回，袋中有$3$条蓝色、$3$条红色和$3$条绿色腕带。他们被分成蓝色组、红色组和绿色组。每组中最高者被任命为组队长。求三组队长是三名最高运动员的概率。

(A) \dfrac{2}{9} \dfrac{2}{9}

(B) \dfrac{2}{7} \dfrac{2}{7}

(D) \dfrac{1}{3} \dfrac{1}{3}

(E) \dfrac{3}{8} \dfrac{3}{8}

Answer

Correct choice: (C)

正确答案：（C）

Solution

We will arrange the nine athletes in a line, with the first three being in the blue group, the next three being in the red group, and the last three being in the green group. Suppose the three tallest athletes are named $A, B, C$, in some order. We have $3!$ ways to choose which group each of these athletes can be in. We then have $3 \cdot 3 \cdot 3$ ways to determine which of the three positions in that group they can take. From here, we must distribute the remaining six athletes in the remaining six spaces, which can be done in $6!$ ways. There are therefore $6 \cdot 27 \cdot 6!$ favorable outcomes. There are also $9!$ total ways to arrange the athletes with no restrictions. Our answer is $\frac{6 \cdot 27 \cdot 6!}{9!} = \frac{6 \cdot 27}{9 \cdot 8 \cdot 7} = \boxed{\textbf{(C) }\frac{9}{28}}.$

我们将九名运动员排成一列，前三名为蓝色组，接下来三名为红色组，最后三名为绿色组。假设三名最高运动员分别为$A, B, C$，顺序任意。有$3!$种方式选择这些运动员分别属于哪个组。然后有$3 \cdot 3 \cdot 3$种方式决定他们在该组中的三个位置。从这里，必须将剩余六名运动员分配到剩余六个位置，有$6!$种方式。因此，有$6 \cdot 27 \cdot 6!$种有利结果。总共有$9!$种无限制排列方式。答案是$\frac{6 \cdot 27 \cdot 6!}{9!} = \frac{6 \cdot 27}{9 \cdot 8 \cdot 7} = \boxed{\textbf{(C) }\frac{9}{28}}$。

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