2025 AMC 10B Problem 11

Six students meet six tutors at the same time, so each day's assignment is a one-to-one matching: a permutation. Monday's matching is irrelevant. Rename the tutors so that on Monday tutor $i$ saw student $i$; then Tuesday's matching is a uniformly random permutation $\sigma$ of $\{1, \dots, 6\}$, and student $i$ sees the same tutor twice exactly when $\sigma(i) = i$.

So the question is: what is the probability that a random permutation of $6$ elements has exactly $2$ fixed points?

Choose the two repeat students in $\binom{6}{2} = 15$ ways. The other four must all move, so they must be permuted with no fixed points, i.e. deranged. Count the derangements of four elements by cycle type:

Two 2-cycles. Pair up the four students: $3$ ways.

One 4-cycle. Fix the smallest element and order the other three after it: $3! = 6$ ways.

There are no other fixed-point-free types, so there are $3 + 6 = 9$ derangements.

Favourable permutations: $15 \cdot 9 = 135$. Total permutations: $6! = 720$. Hence
$$\frac{135}{720} = \frac{3}{16}.$$

The answer is $\boxed{\textbf{(B)}\ \frac{3}{16}}$.

Any "repeat matching" question between two independent random matchings of the same two sets reduces to fixed points of a single random permutation, because composing Tuesday's assignment with the inverse of Monday's is still uniform. That reduction removes one of the two random stages for free. From there, "exactly $m$ fixed points" always factors as $\binom{n}{m}$ times the number of derangements of the remaining $n - m$ objects; the small derangement numbers $D_1 = 0$, $D_2 = 1$, $D_3 = 2$, $D_4 = 9$ are worth knowing outright.

Get $D_4$ by inclusion-exclusion instead of by cycle type: of the $4! = 24$ permutations, subtract those fixing at least one element,
$$24 - \binom{4}{1}3! + \binom{4}{2}2! - \binom{4}{3}1! + \binom{4}{4}0! = 24 - 24 + 12 - 4 + 1 = 9.$$

• Choosing the 2 repeat students and then assigning the other 4 freely, C(6,2)*4!/6! = 1/2 at choice (E), which forgets that the other four must all change tutors.
• Reading the assignment as each student independently picking one of the six tutors; that model gives $\binom{6}{2}(1/6)^2(5/6)^4$, which is not among the choices.
• Using $D_4 = 8$ or $D_4 = 6$ from a misremembered formula; deranging four objects allows both a $4$-cycle and a pair of transpositions.