AMC 10 · 2025 · #11
Grade 7 probabilityOn 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?
Pick an answer.
AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Six students are each randomly matched to one of six tutors on Monday, one student per tutor. On Tuesday the same six students are randomly matched to the same six tutors again, one per tutor. Find the probability that exactly $2$ of the students end up with their Monday tutor a second time.
Givens: $6$ students and $6$ tutors, matched one-to-one each day; Monday: each student is randomly assigned to a tutor; Tuesday: the same students are randomly assigned again, independently; "Met the same tutor both days" means a student's Tuesday tutor equals their Monday tutor; Answer choices: (A) $\tfrac{1}{16}$, (B) $\tfrac{3}{16}$, (C) $\tfrac{1}{4}$, (D) $\tfrac{3}{8}$, (E) $\tfrac{1}{2}$
Unknowns: The probability that exactly $2$ of the $6$ students are paired with their Monday tutor again on Tuesday
Understand
Restated: Six students are each randomly matched to one of six tutors on Monday, one student per tutor. On Tuesday the same six students are randomly matched to the same six tutors again, one per tutor. Find the probability that exactly $2$ of the students end up with their Monday tutor a second time.
Givens: $6$ students and $6$ tutors, matched one-to-one each day; Monday: each student is randomly assigned to a tutor; Tuesday: the same students are randomly assigned again, independently; "Met the same tutor both days" means a student's Tuesday tutor equals their Monday tutor; Answer choices: (A) $\tfrac{1}{16}$, (B) $\tfrac{3}{16}$, (C) $\tfrac{1}{4}$, (D) $\tfrac{3}{8}$, (E) $\tfrac{1}{2}$
Plan
Primary tool: #15 Organize Information in More Ways
Secondary: #2 Make a Systematic List, #16 Change Focus / Count the Complement, #7 Identify Subproblems
Two separate random days sound complicated, so Tool #15 (Organize Information in More Ways) rewrites them as one object: fix Monday as the reference, and Tuesday becomes a single random reshuffle of the students. Then "met the same tutor both days" simply means a student stays put — a fixed point of the reshuffle. Tool #2 (Make a Systematic List) counts the equally likely outcomes: all $6!$ reshuffles, and the $\binom{6}{2}$ choices of which two stay. Tool #16 (Count the Complement) handles the hard part — the other four must all move, an "everybody switches" count called a derangement. Tool #7 (Identify Subproblems) keeps these three counts separate so each stays easy.
Execute — Answer: B
7.SP.C.7 Step 1 Turn two days into one shuffle
- Freeze Monday's matches as the reference picture.
- Tuesday hands the same six students back out to the same six tutors, one each, so Tuesday is just a random reshuffle of the students compared to Monday.
- A student "meets the same tutor both days" exactly when the shuffle drops them back into their Monday tutor — a spot that stays unchanged.
- So the question becomes: in a random reshuffle of $6$ items, what is the chance that exactly $2$ stay in place?
💡 Comparing Tuesday to a frozen Monday turns two random events into one thing you can count.
7.SP.C.8 Step 2 Count all equally likely outcomes
- Count every possible Tuesday matching.
- The first tutor can get any of $6$ students, the next any of the remaining $5$, then $4$, $3$, $2$, $1$.
- Multiplying gives $6! = 720$ equally likely one-to-one matchings.
- This is the denominator of the probability.
💡 Filling tutors one at a time with shrinking choices counts every matching without missing or repeating any.
7.SP.C.8 Step 3 Pick which 2 students stay
- For exactly two students to keep their Monday tutor, first choose which two those are.
- Choosing $2$ students out of $6$, where order does not matter, is $\binom{6}{2} = \tfrac{6\cdot5}{2} = 15$ ways.
💡 Deciding who stays is a simple pick-two choice before worrying about anyone moving.
7.SP.C.8 Step 4 Force the other 4 to all move
- The remaining $4$ students must each land with a different tutor than Monday's, or else more than two would stay.
- Count the reshuffles of $4$ items where nobody stays put (a derangement).
- Start from all $4! = 24$ reshuffles and remove the ones with at least one stayer using inclusion-exclusion: subtract the $\binom{4}{1}3!=24$ with a chosen stayer, add back the $\binom{4}{2}2!=12$ double-counted, subtract $\binom{4}{3}1!=4$, add $\binom{4}{4}0!=1$.
- That leaves $24-24+12-4+1 = 9$ ways where all four move.
💡 Counting "nobody stays" is easiest by starting from everything and peeling away the arrangements that let someone stay.
4.OA.A.3 Step 5 Combine into favorable outcomes
- Each favorable Tuesday is built by first choosing the two stayers ($15$ ways) and then deranging the other four ($9$ ways), and these choices are independent, so multiply.
- That gives $15 \cdot 9 = 135$ matchings with exactly two students back at their Monday tutor.
💡 A two-stage build multiplies its stage counts, since every first choice pairs with every second choice.
7.NS.A.2 Step 6 Divide and simplify
- The probability is favorable over total: $\tfrac{135}{720}$.
- Both share a factor of $45$, and $135 = 45\cdot 3$ while $720 = 45\cdot 16$, so the fraction reduces to $\tfrac{3}{16}$.
- That matches choice (B).
💡 Probability is just the good outcomes over all outcomes, then reduced to lowest terms.
7.SP.C.7 Freeze Monday's matches as the reference picture. Tuesday hands the same six stu 7.SP.C.8 Count every possible Tuesday matching. The first tutor can get any of $6$ studen 7.SP.C.8 For exactly two students to keep their Monday tutor, first choose which two thos 7.SP.C.8 The remaining $4$ students must each land with a different tutor than Monday's, 4.OA.A.3 Each favorable Tuesday is built by first choosing the two stayers ($15$ ways) an 7.NS.A.2 The probability is favorable over total: $\tfrac{135}{720}$. Both share a factor Review
Reasonableness: The number of favorable outcomes, $135$, is less than the total $720$, so the probability is under $1$ — sanity passes. It should also be small, since needing exactly two returns (no more, no fewer) is a fairly special event, and $\tfrac{3}{16}\approx 0.19$ fits that. A cross-check: the fraction of all six-item reshuffles with a given number of fixed points, using $\binom{6}{k}D_{6-k}$, gives counts $265,264,135,40,15,0,1$ for $k=0,1,2,3,4,5,6$, which sum to $720$ exactly — and the $k=2$ entry is our $135$. This confirms both the count and that no outcomes were double-counted, so $\tfrac{3}{16}$ is right.
Alternative: Instead of choosing the stayers first, build the probability student by student. The chance a specific set of $2$ students returns and the other $4$ all move can be computed with conditional fractions, then multiplied by the $\binom{6}{2}$ ways to pick the pair; this yields the same $\tfrac{15\cdot 9}{720}=\tfrac{3}{16}$. The counting method shown is cleaner because it keeps whole-number counts until the final divide, avoiding messy fraction chains.
CCSS standards used (min grade 7)
7.SP.C.7Develop probability models and use them to find probabilities of events (Reframing the two random days as one random reshuffle and setting up the probability as favorable matchings over all matchings.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Counting the $6!=720$ total matchings, the $\binom{6}{2}=15$ ways to choose the two stayers, and the $9$ derangements of the remaining four.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Multiplying the stage counts $15\cdot 9$ to get $135$ favorable outcomes.)7.NS.A.2Apply and extend understanding of multiplication and division of rational numbers (Dividing $135$ by $720$ and reducing the fraction to $\tfrac{3}{16}$.)
⭐ Freeze the first day and treat the second as a reshuffle: pick which $2$ students stay ($15$ ways), make the other $4$ all move ($9$ ways), and divide $15\cdot 9=135$ by $720$ to get $\tfrac{3}{16}$.
⭐ Freeze the first day and treat the second as a reshuffle: pick which $2$ students stay ($15$ ways), make the other $4$ all move ($9$ ways), and divide $15\cdot 9=135$ by $720$ to get $\tfrac{3}{16}$.
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