AMC10 2025 A

AMC10 2025 A · Q14

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

Six chairs are arranged around a round table. Two students and two teachers randomly select four of the chairs to sit in. What is the probability that the two students will sit in two adjacent chairs and the two teachers will also sit in two adjacent chairs?

六把椅子围成圆桌排列。两名学生和两名老师随机选四把椅子坐下。两名学生坐在相邻两把椅子的概率，且两名老师也坐在相邻两把椅子的概率是多少？

(A) \frac 16 \frac 16

(B) \frac 15 \frac 15

(D) \frac 3{13} \frac 3{13}

(E) \frac 14 \frac 14

Answer

Correct choice: (B)

正确答案：（B）

Solution

Pair two students together and put them adjacent on any two seats. There are 6 ways to do this. Considering one of these cases (they are all the same), there are 4 seats left, in which we wish to arrange the teachers together. So pair the teachers together and put them adjacent on any two seats not already occupied by two of the students. There are 3 ways to do this. For all 6 cases, there are 6×3=18 favorable outcomes. The number of ways to arrange the 2 students and 2 teachers is $\binom{6}{2} \times \binom{4}{2} = 90$. Our probability is 1890= $\boxed{\textbf{(B) } \frac 15}$

将两名学生配对放在相邻的任意两把座位上。有 $6$ 种方法。考虑其中一种情况（它们都相同），剩下 $4$ 把座位，我们希望老师们也坐在一起。所以将老师们配对放在未被学生占据的任意相邻两把座位上。有 $3$ 种方法。对于所有 $6$ 种情况，有 $6\times3=18$ 种有利结果。排列 $2$ 名学生和 $2$ 名老师的方法数为 $\binom{6}{2} \times \binom{4}{2} = 90$。概率为 $\frac{18}{90}= \boxed{\textbf{(B) } \frac 15}$

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