M32-05

Official Solution:

At a birthday party, 10 students are to be seated around a circular table. Two seating arrangements are considered different only when the positions of the people are different relative to each other. What is the probability that two of the students, Anna and Bill, do NOT sit next to each other?

A. \(\frac{8}{9}\)
B. \(\frac{7}{9}\)
C. \(\frac{4}{5}\)
D. \(\frac{2}{9}\)
E. \(\frac{1}{9}\)


There are 10 students around a circular table, which can be arranged in \((10-1)!=9!\) ways.

Now, consider Anna and Bill as a single unit - {Anna, Bill}. Then, we have nine units to arrange, namely eight students and {Anna, Bill}. These units can be arranged around the circular table in \((9-1)!=8!\) ways. Anna and Bill within their unit can be arranged in two ways {Anna, Bill} or {Bill, Anna}. Thus, the total number of ways to arrange the students such that Anna and Bill sit next to each other is \(8!*2\). Therefore, the number of ways to arrange the students so that Anna and Bill do NOT sit next to each other is \(9!-8!*2\).

The probability of Anna and Bill not sitting next to each other is then \(\frac{favorable}{total}=\frac{9!-8!*2}{9!}=1-\frac{2}{9}=\frac{7}{9}\).


Answer: B