$36 \cdot 24$ Nine chairs are arranged in a row, and there are 6 students and three professors (denoted as $\alpha$, $\beta$, and $\gamma$). The three professors arrive first and decide to choose their seats so that each professor is seated between two students. The number of such seating arrangements is
(A) 12.
(B) 36.
(C) 60.
(D) 80.
(E) 630.
(45th American High School Mathematics Examination, 1994)
[Solution] 6 students standing in a row have 5 gaps, and three professors $\alpha, \beta, \gamma$ each choose one of them. The number of ways to arrange them is
$$
P_{5}^{3}=5 \times 4 \times 3=60 \text{ (ways) }.
$$

Therefore, the answer is $(C)$.