AMC 10 · 2025 · #14
Grade 7 probabilityPick an answer.
AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Count every possible seating
Seat four distinct people in six distinct chairs one at a time: 6, then 5, then 4, then 3 choices, for 360 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.
Each person who sits uses up one chair, so the open chairs drop by one for the next person.
▸ Why?
Each stage is a free choice among the chairs still open, so the counts multiply.
▸ Why?
Each person takes exactly one chair, so people and chairs pair off without leftovers.
Stage A: seat the students together
A ring of 6 chairs holds 6 neighboring pairs; pick one for the students and decide who sits where, giving 12 ways.
A ring of 6 chairs has exactly 6 neighbor-pairs, one starting at each chair.
7.SP.C.8Identify SubproblemsStage B: seat the teachers together
The students' block breaks the ring into a line of 4 chairs, which holds only 3 neighboring pairs, so the teachers have 6 ways.
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.8Draw A DiagramDivide favorable by total
Any student pair pairs with any teacher pair, so 72 favorable seatings out of 360 reduce to 1/5.
Favorable outcomes over all equally likely outcomes is exactly what probability means.
7.SP.C.7Identify SubproblemsSeat 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.
- Count every possible seating
- Stage A: seat the students together
- Stage B: seat the teachers together
- Divide favorable by total