2025 AMC 10A Problem 14

P11-15~4 minPrint

Basic Probability Arrangements with Restrictions Set up the equation/formula and compute; no special trick needed Exploit symmetry to reduce work or pair up objects

Six 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?

Show solution without answering

Key insight

The students sit side by side with chance 6/15. The four chairs left then form a row, not a circle, so the teachers have 3 good pairs out of 6.

Solution

Number the chairs $1$ to $6$ around the table, so chair $6$ is next to chair $1$ .

Whether two people are side by side depends only on which two chairs they take, not on who sits in which one. So think of it this way: the students pick a pair of chairs, then the teachers pick a pair from the chairs left over. Every pair is equally likely at each step.

Step 1: the students.

The number of ways to pick $2$ chairs out of $6$ is $\binom{6}{2} = 15$ .
The side-by-side pairs are $\{1,2\}, \{2,3\}, \{3,4\}, \{4,5\}, \{5,6\}, \{6,1\}$ . That is $6$ pairs (do not forget $\{6,1\}$ , which wraps around the table).
So the chance the students sit together is $\frac{6}{15} = \frac25$ .

Step 2: the teachers, once the students are together.

Every side-by-side pair looks the same around a round table, so suppose the students are in chairs $1$ and $2$ . The chairs left are $3, 4, 5, 6$ .
The teachers can pick $\binom{4}{2} = 6$ pairs from these four chairs.
Which of those pairs are side by side? $\{3,4\}$ , $\{4,5\}$ and $\{5,6\}$ . What about $\{6,3\}$ ? Chairs $6$ and $3$ are not neighbors, because chairs $1$ and $2$ sit between them. The four empty chairs form a row, not a circle.
So the chance the teachers sit together is $\frac{3}{6} = \frac12$ .

Step 3: combine. Both things must happen, so multiply:
$\frac25 \cdot \frac12 = \frac15.$

The answer is $\boxed{\textbf{(B)}\ \frac{1}{5}}$ .

Why this works

Counting pairs of chairs instead of seatings keeps the numbers small, and it is safe because swapping the two students (or the two teachers) never changes who is next to whom. The key moment is Step 2: taking two neighboring chairs out of a circle leaves a row, and a row of $4$ chairs has only $3$ neighboring pairs, while a circle of $4$ would have $4$ . Always check whether what is left is still a circle.

Alternative approach

Count everything directly instead of multiplying chances.

All outcomes: the students pick one of $15$ pairs, then the teachers pick one of the $6$ pairs from the four chairs left. That is $15 \cdot 6 = 90$ outcomes, all equally likely.
Good outcomes: the students pick one of the $6$ side-by-side pairs. For each one, Step 2 above showed the teachers have $3$ side-by-side pairs left. That is $6 \cdot 3 = 18$ good outcomes.

So the probability is $\frac{18}{90} = \frac15$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Only which pair of chairs each group occupies matters, not who sits where inside a pair. So think of it as the students choosing $2$ of the $6$ chairs, then the teachers choosing $2$ of the remaining $4$ ; every one of the $\binom{6}{2}\binom{4}{2} = 15 \cdot 6 = 90$ outcomes is equally likely.

Students adjacent. Around a circle of $6$ chairs there are exactly $6$ adjacent pairs, so this has probability $\dfrac{6}{15} = \dfrac{2}{5}$ .

Teachers adjacent, given that. Removing two neighbouring chairs from the circle leaves four chairs in a row: for students in chairs $1$ and $2$ , the chairs $3, 4, 5, 6$ remain, and chairs $3$ and $6$ are not neighbours, since $1$ and $2$ sit between them. Of the $\binom{4}{2} = 6$ pairs available to the teachers, the adjacent ones are $\{3,4\}, \{4,5\}, \{5,6\}$ , so the probability is $\dfrac{3}{6} = \dfrac12$ .

Multiplying,
$P = \frac{2}{5}\cdot\frac{1}{2} = \frac{1}{5}.$

The answer is $\boxed{\textbf{(B)}\ \frac{1}{5}}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.5)
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques