2025 AMC 12A Problems/Problem 6

Problem

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?

$\textbf{(A) } \frac 16 \qquad \textbf{(B) } \frac 15 \qquad \textbf{(C) } \frac 29 \qquad \textbf{(D) } \frac 3{13} \qquad \textbf{(E) } \frac 14$

Solution 1

We first count the number of ways to place $4$ distinct people into $6$ distinct chairs: $6\cdot5\cdot4\cdot3 = 360$ .

We now count how many favorable assignments there are. There are $6$ ways to choose an adjacent pair of chairs for the two students. We can arrange the two students in those two chairs in $2$ ways.

After those two adjacent chairs are taken, there are $4$ chairs left, with $3$ adjacent pairs among them. We choose one of these pairs, arranging the teachers for $3 \cdot 2 = 6$ ways.

There are $6\cdot 2 \cdot 6 = 72$ favorable arrangements.

The probability is therefore $\frac{72}{360} = \boxed{\frac{1}{5}}$

~lprado

Solution 2

We split the problem into cases of where the first teacher and first student sits. A student sits in a seat with $1$ probability

Case 1: The first teacher sits next to the first student. This will happen with 2/5 probability. Now, there is one valid seat for the second student to sit in with probability 1/4 and and one valid valid seat for the second teacher to sit in with probability 1/3. Total probability of this case is 2/5 * 1/4 * 1/3 = 1/30

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2025 AMC 12A Problems/Problem 6

Problem

Solution 1

Solution 2