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20 Jun 2016, 04:30
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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. \(\frac{1}{9}\)
A. \(\frac{8}{9}\)
B. \(\frac{7}{9}\)
C. \(\frac{4}{5}\)
D. \(\frac{2}{9}\)
E. \(\frac{1}{9}\)
20 Jun 2016, 04:30
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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
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
07 Aug 2016, 23:13
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manlog
I solved it in a bit simpler form:
There are 10 spots for 10 people. How can we break the requirement? What's the probability of arranging these two people in a way that they sit next to each other?
Probability of choosing a person is 1, then what's the probability of choosing a person sitting next to him = 1/9 * 2 (since there are two seats right next to him). Then the probability of arranging these two people in a way, that they don't seat to each other is 1 - 2/9 = 7/9
There are 10 spots for 10 people. How can we break the requirement? What's the probability of arranging these two people in a way that they sit next to each other?
Probability of choosing a person is 1, then what's the probability of choosing a person sitting next to him = 1/9 * 2 (since there are two seats right next to him). Then the probability of arranging these two people in a way, that they don't seat to each other is 1 - 2/9 = 7/9
My approach is very similar -
1. Let's say Ann sits at a place.
2. Probability that Bill sits next to Ann in circular arrangement is \(\frac{[Left + Right]}{[Remaining 9 places]}\) = \(\frac{2}{9}\)
3. Probability of both not sitting together is \(1 - \frac{2}{9} = 7/9\)
