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At a birthday party, 10 students are to be seated around a circular

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At a birthday party, 10 students are to be seated around a circular  [#permalink]

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New post 19 Nov 2019, 03:30
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At a birthday party, 10 students are to be seated around a circular table. What is the probability that two of the students, Anna and Bill, do NOT sit next to each other?

A) 8/9
B) 7/9
C) 5/9
D) 2/9
E) 1/9

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At a birthday party, 10 students are to be seated around a circular  [#permalink]

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New post Updated on: 19 Nov 2019, 04:04
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Raxit85 wrote:
At a birthday party, 10 students are to be seated around a circular table. What is the probability that two of the students, Anna and Bill, do NOT sit next to each other?

A) 8/9
B) 7/9
C) 5/9
D) 2/9
E) 1/9


Ans. is B

Simple Approach -

1. Let's say Ann sits at a place.
2. Probability that Bill sits next to Ann in circular arrangement is [Left+Right]/[Remaining 9 places]= 2/9
3. Probability of both not sitting together is 1−2/9=7/9

Also, Pls Don't post same questions, which are already available on GmatClub Site.
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Pls provide kudos, if my explanation is good enough.

Originally posted by rajatchopra1994 on 19 Nov 2019, 03:45.
Last edited by rajatchopra1994 on 19 Nov 2019, 04:04, edited 1 time in total.
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Re: At a birthday party, 10 students are to be seated around a circular  [#permalink]

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New post 19 Nov 2019, 03:49
Raxit85 wrote:
At a birthday party, 10 students are to be seated around a circular table. What is the probability that two of the students, Anna and Bill, do NOT sit next to each other?

A) 8/9
B) 7/9
C) 5/9
D) 2/9
E) 1/9


Official Explanation



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

Now, consider Anna and Bill as one unit - {Anna, Bill}. We will have 9 units to arrange: 8 students and {Anna, Bill}. Those 9 units can be arranged around a 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 number of ways to arrange 10 students around a circular table so that two of them, Ana and Bill, sit next to each other is \(8!*2\).

Therefore, the number of ways to arrange 10 students around a circular table so that two of them, Ana and Bill, do NOT sit next to each other is \(9!-8!*2\)

The probability = \(\frac{favorable}{total}=\frac{9!-8!*2}{9!}=1-\frac{2}{9}=\frac{7}{9}\)


Answer: B
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Re: At a birthday party, 10 students are to be seated around a circular  [#permalink]

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New post 22 Nov 2019, 12:57
Raxit85 wrote:
At a birthday party, 10 students are to be seated around a circular table. What is the probability that two of the students, Anna and Bill, do NOT sit next to each other?

A) 8/9
B) 7/9
C) 5/9
D) 2/9
E) 1/9

M32-05



The number of ways for Anna and Bill to sit next to each other is:

2! * (9 - 1)! = 2 * 8!

The total number of ways to arrange the 10 people in a circle is (10 - 1)! = 9!

The probability that Anna and Bill will sit together is (2 * 8!)/9! = 2/9, so the probability that they won’t sit together is 7/9.

Answer: B
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Re: At a birthday party, 10 students are to be seated around a circular   [#permalink] 22 Nov 2019, 12:57
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