My attempt:
If no 2 people can sit next to each other then I can take 1 people and 2 chairs as 1 unit. So the first person has 15 choices where to sit, second has 14, third has 13 and so on.. Until the eighth person has 15-7 choices.
$${15! \over (15-8)!} = {15! \over 7!}$$
Am I doing it right? I have a feeling that there is something more to it..
Best Answer
Line up the eight people in some order, say alphabetically. Hand each of them a chair. That leaves $22$ chairs. Line up those $22$ chairs in a row. This creates $23$ spaces, $21$ between successive chairs and $2$ at the ends of the row, in which a person can insert a chair. To ensure that the people are separated, each person must place his or her chair in a separate space. The first person has $23$ options, which leaves $22$ options for the next person, and so forth. Hence, the number of permissible seating arrangements is $$23 \cdot 22 \cdot 21 \cdot 20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 = 23 \cdot 22 \cdot 21 \cdot 20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 \cdot \frac{15!}{15!} = \frac{23!}{15!}$$