The 2 adjacent students will have an arrangement of 5. Next, the open 4 chairs can be filled by either of the 3 students for 3!= 6. The product of 5 and 6 = 30. Am I taking the proper approach?
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You can think of it is how many ways can 5 things (the pair, the other 3 students, and the empty seat) be arranged. And then for each of those arrangements, there are multiple ways the pair can be arranged, so multiply by that number.
turkeyhundt
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You are not ordering the couple and you are not taking into account that there are $4$ available seats after seating the two people that must be together. The first part is good.
There are $5$ ways to select the two adjacent seats for the two adjacent students and then $2$ ways to select which one sits in which of the two seats. After this there are $4$ empty seats for $3$ people. There are $4$ ways to pick the empty seat, and after this there are indeed $3!$ ways to distribute the $3$ students in the $3$ selected seats.
hence the final answer is $5\cdot 2\cdot 4\cdot 3!=40\cdot 6=240$
Asinomás
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