AMC10 2025 B

AMC10 2025 B · Q11

AMC10 2025 B · Q11. It mainly tests Combinations, Counting & probability misc.

On Monday, $6$ students went to the tutoring center at the same time, and each one was randomly assigned to one of the $6$ tutors on duty. On Tuesday, the same $6$ students showed up, the same $6$ tutors were on duty, and the students were again randomly assigned to the tutors. What is the probability that exactly $2$ students met with the same tutor both Monday and Tuesday?

周一，有$6$名学生同时来到辅导中心，每人被随机分配到值班的$6$名辅导老师中的一位。周二，这$6$名学生再次出现，相同的$6$名老师值班，学生们再次被随机分配到老师那里。求恰好有$2$名学生周一和周二都遇到同一名老师的概率。

(A) \ \frac{1}{16} \ \frac{1}{16}

(B) \ \frac{3}{16} \ \frac{3}{16}

(D) \ \frac{3}{8} \ \frac{3}{8}

(E) \ \frac{1}{2} \ \frac{1}{2}

Answer

Correct choice: (B)

正确答案：（B）

Solution

On Monday, the first tutor has 6 choices for kids to be assigned to, the second has 5, and so on. Hence, there are $6!$ possible ways to assign the tutors to the kids. On Tuesday, there are $\binom{6}{2}$ ways to choose the kid-tutor pairs that remain the same from Monday. For the other 4 kids, they need to be assigned to the other tutors in such a way that none of those kids are assigned to the same tutor that they were assigned to on Monday. There are $9$ ways to do this. Thus, our final answer is $\frac{\binom{6}{2} \cdot 9}{6!}$ or $\boxed{\textbf{(B)}\hspace{3pt}\frac{3}{16}}$. Remark: We can get $9$ either by listing all the possible configurations or by rounding $\frac{4!}{e}$. This is called a derangement and is notated as $!n$. It can be calculated with the explicit formula $!n = n!\sum _{i=0}^{n}\frac{(-1)^{i}}{i!}$

周一，第一位老师有$6$种学生选择，第二位有$5$种，以此类推。因此，分配学生给老师共有$6!$种可能方式。周二，有$\binom{6}{2}$种方式选择周一和周二保持相同的学生-老师配对。对于其余$4$名学生，他们需要被分配到其他老师那里，使得这些学生中没有一人被分配到周一时的同一老师。有$9$种方式做到这一点。因此，最终答案是$\frac{\binom{6}{2} \cdot 9}{6!}$，即$\boxed{\textbf{(B)}\hspace{3pt}\frac{3}{16}}$。备注：$9$可以通过列举所有可能配置得到，或者通过取整$\frac{4!}{e}$得到。这称为错位排列，记作$!n$。其显式公式为$!n = n!\sum _{i=0}^{n}\frac{(-1)^{i}}{i!}$。

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