2025 AMC 12A Problems/Problem 6

Problem

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

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2025 AMC 12A Problems/Problem 6

Problem

Solution 1