AMC12 2025 A

AMC12 2025 A · Q6

AMC12 2025 A · Q6. 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?

六把椅子围成一圈摆放。两名学生和两名老师随机选择四把椅子坐下。两名学生坐在相邻的两把椅子上且两名老师也坐在相邻的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任意两把相邻的座位上，有6种方式。考虑其中一种情况（它们都相同），剩下4把座位，我们希望将老师们安排在一起。所以将老师们配对放在任意两把未被学生占据的相邻座位上，有3种方式。对于所有6种情况，有6×3=18种有利结果。安排2名学生和2名老师的总数是$\binom{6}{2} \times \binom{4}{2} = 90$。概率是$\frac{18}{90}=\boxed{\textbf{(B) } \frac{1}{5}}$

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