1994 AHSME Problems/Problem 22

Problem

Nine chairs in a row are to be occupied by six students and Professors Alpha, Beta and Gamma. These three professors arrive before the six students and decide to choose their chairs so that each professor will be between two students. In how many ways can Professors Alpha, Beta and Gamma choose their chairs?

$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 36 \qquad\textbf{(C)}\ 60 \qquad\textbf{(D)}\ 84 \qquad\textbf{(E)}\ 630$

Solution

Let the seats be numbered in order clockwise around the table $1,2,3,\dots 9$ . Let $a,b,c$ be the seats occupied by Alpha, Beta, and Gamma, respectively. Without loss of generality, fix Alpha at seat $1$ to account for rotations. Then there are ten choices for the set $\{b,c\}$ , namely $\{b,c\}=\{3,5\},\{3,6\},\{3,7\},\{3,8\},\{4,6\},\{4,7\},\{4,8\},\{5,7\},\{5,8\},\{6,8\}$ . After multiplying by $3!=6$ (number of ways to permute Alpha, Beta, and Gamma), we get the final answer of $\boxed{\textbf{(C)}\ 60}$

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1994 AHSME Problems/Problem 22

Problem

Solution