2025 AMC 10B Problem 11

1. Fix Monday. Whatever happened on Monday, each student now has a "Monday tutor". Only Tuesday is random. There are $6! = 720$ equally likely ways to match $6$ students with $6$ tutors on Tuesday.

1. Plan the count. We want exactly $2$ students to get their Monday tutor again. First choose which $2$ students repeat, then make sure the other $4$ all get a new tutor.

1. Choose the two repeaters. There are $\binom{6}{2} = 15$ ways.

4. The other four must all change. Call these students $A, B, C, D$ and their Monday tutors $a, b, c, d$. The four tutors $a, b, c, d$ are left for them, and nobody may get their own letter.
- $A$ can get $b$, $c$ or $d$: $3$ choices. The three cases look alike, so work out $A \to b$.
- Then $B$ can get $a$, $c$ or $d$.
- $B \to a$: $C$ and $D$ must swap, $C \to d$, $D \to c$. $1$ way.
- $B \to c$: tutors $a, d$ are left. $D$ cannot get $d$, so $D \to a$ and $C \to d$. $1$ way.
- $B \to d$: tutors $a, c$ are left. $C$ cannot get $c$, so $C \to a$ and $D \to c$. $1$ way.
- That is $3$ ways for each choice for $A$, so $3 \times 3 = 9$ ways.

5. Put it together. Good Tuesday matchings: $15 \times 9 = 135$. So the probability is
$$\frac{135}{720} = \frac{3}{16}.$$

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

With two random days, fix the first one and only the second matters, which halves the work. "Exactly $2$ repeat" means two things: the chosen $2$ repeat and nobody else does. Forgetting the second part is the usual mistake. The numbers of ways to mix up $2$, $3$ or $4$ items so none stays in place are $1$, $2$ and $9$, and they are worth remembering.

Count the ways for the four students to all change by taking away the bad cases. Of the $4! = 24$ ways to hand out $a, b, c, d$:
- exactly $1$ student gets their own tutor: choose that student ($4$ ways), and the other three all change ($2$ ways), so $8$;
- exactly $2$ get their own: choose them ($6$ ways), and the other two swap ($1$ way), so $6$;
- exactly $3$ get their own is impossible, since the fourth would then have to as well;
- all $4$ get their own: $1$.

That leaves $24 - 8 - 6 - 1 = 9$ ways, as before.

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