AMC 10 · 2025 · #14
Grade 7 probabilitySix chairs are arranged around a round table. Two students and two teachers randomly select four of the chairs to sit in. What is the probability that the two students will sit in two adjacent chairs and the two teachers will also sit in two adjacent chairs?
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 chairs sit in a ring around a round table. Two students and two teachers each pick one of the chairs, so four of the six chairs get used. Find the probability that the two students end up next to each other and the two teachers also end up next to each other.
Givens: There are $6$ chairs spaced evenly around a round table, so each chair is a fixed, distinct spot; Two students and two teachers each choose a chair at random; Exactly $4$ of the $6$ chairs are filled, one person per chair
Unknowns: The probability that the two students are in adjacent chairs and the two teachers are also in adjacent chairs
Understand
Restated: Six chairs sit in a ring around a round table. Two students and two teachers each pick one of the chairs, so four of the six chairs get used. Find the probability that the two students end up next to each other and the two teachers also end up next to each other.
Givens: There are $6$ chairs spaced evenly around a round table, so each chair is a fixed, distinct spot; Two students and two teachers each choose a chair at random; Exactly $4$ of the $6$ chairs are filled, one person per chair
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List, #1 Draw a Diagram
Probability here is (favorable seatings) divided by (all seatings), so the job is two counts. Tool #7 (Identify Subproblems) splits the favorable count into two clean stages: first seat the students as an adjacent pair, then seat the teachers as an adjacent pair in the chairs that are left. Tool #2 (Make a Systematic List) counts how many adjacent pairs a circle of $6$ chairs has, and how many ways two named people fill a chosen pair. Tool #1 (Draw a Diagram) reveals the key twist in the second stage: once the students take one adjacent pair out of the ring, the $4$ chairs that remain form a broken arc (a line), not a full circle, so they have fewer adjacent pairs than you might expect.
Execute — Answer: B
4.OA.A.3 Step 1 Count every possible seating
- The four people are distinct and the six chairs are distinct fixed spots, so seat them one at a time.
- The first person has $6$ chairs to choose from, the next has $5$ left, then $4$, then $3$.
- Multiplying these choices gives the total number of equally likely seatings.
💡 Each person that sits down uses up one chair, so the number of open chairs drops by one for the next person.
7.SP.C.8 Step 2 Stage A: seat the students together
- Break the favorable count into two stages and handle the students first.
- Going around the ring of $6$ chairs, the pairs of neighbors are $(1,2),(2,3),(3,4),(4,5),(5,6),(6,1)$, which is $6$ adjacent pairs.
- Pick one such pair for the students, then decide which student takes which of the two chairs, giving $2$ orders.
- So the students can be seated as a neighboring pair in $6 \times 2$ ways.
💡 A ring of $6$ chairs has exactly $6$ neighbor-pairs, one starting at each chair.
7.SP.C.8 Step 3 Stage B: seat the teachers together
- Now seat the teachers as a neighboring pair using only the $4$ chairs the students left empty.
- Here is the twist: the students removed one solid block of two neighboring chairs from the ring, so the $4$ leftover chairs no longer close into a circle — they form a single line of $4$ chairs in a row.
- A row of $4$ chairs has neighbor-pairs $(1,2),(2,3),(3,4)$, which is only $3$ adjacent pairs, not $4$.
- Pick one of these $3$ pairs for the teachers and choose which teacher sits where, giving $2$ orders.
💡 Cutting a neighboring block out of a ring turns the rest into a line, and a line of $4$ has one fewer neighbor-pair than a ring of $4$.
7.SP.C.7 Step 4 Divide favorable by total
- By the two stages, the number of favorable seatings is Stage A times Stage B, since any student pair can be combined with any teacher pair.
- Then the probability is that favorable count over the $360$ total seatings, reduced to lowest terms.
- This gives the final answer $\textbf{(B)}\ \tfrac{1}{5}$.
💡 Favorable outcomes over all equally likely outcomes is exactly what probability means.
4.OA.A.3 The four people are distinct and the six chairs are distinct fixed spots, so sea 7.SP.C.8 Break the favorable count into two stages and handle the students first. Going a 7.SP.C.8 Now seat the teachers as a neighboring pair using only the $4$ chairs the studen 7.SP.C.7 By the two stages, the number of favorable seatings is Stage A times Stage B, si Review
Reasonableness: The answer $\tfrac{1}{5}=0.2$ is a probability between $0$ and $1$, so it is at least plausible. A quick recount with unlabeled seats agrees: choose which $4$ of $6$ chairs get used in $\binom{6}{4}=15$ ways, then pick which $2$ of those $4$ belong to the students in $\binom{4}{2}=6$ ways, giving $90$ equally likely patterns. Favorable ones are $6$ student neighbor-pairs times $3$ teacher neighbor-pairs in the remaining line $=18$, and $18/90=\tfrac15$, matching.
Alternative: Instead of seating people, first choose which pair of empty chairs is left unused. Two students plus two teachers fill $4$ chairs, so $2$ chairs stay empty. Count over which two chairs are empty: for the students and teachers to each be a neighboring pair, the used chairs must split the ring into a student-block and a teacher-block. Enumerating these configurations and dividing by all $\binom{6}{2}=15$ empty-pair choices (with the internal orderings) again yields $\tfrac15$.
CCSS standards used (min grade 7)
4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Multiplying the shrinking chair choices $6 \times 5 \times 4 \times 3$ to count all $360$ seatings, and $12 \times 6$ for the favorable ones.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Counting favorable seatings in two stages: adjacent student pairs on the ring, then adjacent teacher pairs on the leftover line of chairs.)7.SP.C.7Develop probability models and use them to find probabilities of events (Dividing the $72$ favorable seatings by the $360$ equally likely total seatings to get the probability $\tfrac15$.)
⭐ Seat the students as a neighbor-pair first, and remember that pulling that block out of the ring leaves the other chairs in a line, which has one fewer neighbor-pair than a circle.
⭐ Seat the students as a neighbor-pair first, and remember that pulling that block out of the ring leaves the other chairs in a line, which has one fewer neighbor-pair than a circle.
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