AMC 10 · 2025 · #6

Grade 7 probability

Archetype Branching State Enumeration └ Sub-archetype Ordered Placement Count

probability-basiccombinations-basicfundamental-counting-principle systematic-enumeration ↑ Prerequisites: combinations-basic

📏 Medium solution 💡 2 insights

Problem

Six labeled chairs sit around a round table. Two students and two teachers each take one of four of the chairs. Find the probability that the two students end up in two neighboring chairs and, at the same time, the two teachers also end up in two neighboring chairs.

Pick an answer.

(A)

$\frac 16$

(B)

$\frac 15$

(C)

$\frac 29$

(D)

$\frac 3{13}$

(E)

$\frac 14$

How to solve

Strategy Make a Systematic List

The question asks what fraction of all equally likely seatings work, so it is a counting problem: count the seatings that work and divide by all seatings. Drawing the ring of six chairs makes 'adjacent' concrete, and splitting the count into 'pick the students' pair' and then 'pick the teachers' pair' keeps the list organized. Because the numbers are small, a careful organized count is faster and safer than any formula, and the five answer choices let me confirm the final fraction lands on a listed value.

1STEP 1

Set up favorable over total

Draw six chairs as a ring. A seating is set by which 2 chairs hold students and which 2 hold teachers; probability is favorable over all.

P = (seatings that work)/(all seatings)

2STEP 2

Count all seatings

Choose 2 of 6 chairs for the students (15 pairs), then 2 of the remaining 4 for the teachers (6 pairs): 15 × 6 = 90 total seatings.

(6 · 5)/2·(4 · 3)/2 = 15 · 6 = 90

3STEP 3

Count where students are adjacent

Around the ring the neighboring pairs are (1,2),(2,3),(3,4),(4,5),(5,6),(6,1) — exactly 6 adjacent pairs out of 15.

(1,2),(2,3),(3,4),(4,5),(5,6),(6,1) → 6 adjacent pairs

4STEP 4

Count teachers adjacent too

With students in chairs 1,2, the open seats 3,4,5,6 form a line: adjacent pairs (3,4),(4,5),(5,6) are 3 each, so 6 × 3 = 18 seatings work.

open seats 3,4,5,6: (3,4),(4,5),(5,6)→ 3; 6 · 3 = 18

5STEP 5

Divide to get the probability

Divide favorable by total: $\frac{18}{90}$ = $\frac{1}{5}$ , which is choice (B). Miscounting teacher pairs as 6 gives $\frac{2}{5}$ — not even listed.

P = 18/90 = 1/5

Answer

1/5

The result

\frac{18}{90}

\frac{1}{5}

= 0.2 lies between 0 and 1, as every probability must, and a one-in-five chance fits a fairly special seating pattern. A different route gives the same value: the chance the two teachers alone land in adjacent chairs is

\frac{6}{15}

\frac{2}{5}

; once they do, the four remaining chairs sit in a row, so 3 of the 6 possible student pairs are adjacent, a chance of

\frac{1}{2}

. Multiplying,

\frac{2}{5}

times

\frac{1}{2}

\frac{1}{5}

, confirming the count.

💡Key takeaway

To find a probability, count the seatings you want and the seatings in all, then divide, and remember that a round table's leftover seats form a line, so they lose one neighbor-pair.

Set up favorable over total
Count all seatings
Count where students are adjacent
Count teachers adjacent too
Divide to get the probability

Set up favorable over total

Count all seatings

Count where students are adjacent

Count teachers adjacent too

Divide to get the probability

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