๐Ÿ”ฅ New! 2025 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 11 Easy

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?

  • A.

    \frac { 1 } { 1 6 }

  • B.

    \frac { 3 } { 1 6 }

  • C.

    \frac { 1 } { 4 }

  • D.

    \frac { 3 } { 8 }

  • E.

    \frac { 1 } { 2 }

Answer:B
Solution

For the general case with n students and n tutors:

Let S_i be the case where the i-th student has the same tutor, then |S_i| = (n-1)! and |S_{i_1} \cap S_{i_2} \cap \cdots \cap S_{i_k}| = (n-k)!.

|(S_1 \cup \cdots \cup S_n)^c| = n! - |S_1 \cup \cdots \cup S_n|

= n! - \left( \sum_{i=1}^{n} |S_i| - \sum_{1 \leq i_1 < i_2 \leq n} |S_{i_1} \cap S_{i_2}| + \cdots + (-1)^{n+1} |S_1 \cap \cdots \cap S_n| \right)

= n! - \left( n \times (n-1)! - \binom{n}{2} \cdot (n-2)! + \cdots + (-1)^{n+1} \cdot 1 \right)

The k-th term inside the bracket equals (-1)^{k+1} \cdot \binom{n}{k} \cdot (n-k)! = (-1)^{k+1} \cdot \frac{n!}{k!}.

This gives n! - \sum_{k=1}^{n} (-1)^{k+1} \cdot \frac{n!}{k!} = n! \cdot \left( \sum_{k=0}^{n} (-1)^k \cdot \frac{1}{k!} \right)

For any 2 students, the total ways of assigning so that they get the same tutor , are the total ways of assigning the remaining 4 tutors so no students among these 4 get the same tutor:

4! \left( 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} \right) = 9

Therefore, the probability = \frac{\binom{6}{2} \times 9}{6!} = \frac{3}{16}

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