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19 Nov 2019, 03:30
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B
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Difficulty:
55%
(hard)
Question Stats:
59% (01:28) correct
41%
(01:35)
wrong
based on 449
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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
M32-05
A) 8/9
B) 7/9
C) 5/9
D) 2/9
E) 1/9
M32-05
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19 Nov 2019, 03:49
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Raxit85
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
Updated on: 19 Nov 2019, 04:04
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.
Last edited by rajatchopra1994 on 19 Nov 2019, 04:04, edited 1 time in total.
Kudos
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Raxit85
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.
Bunuel bb
Pls provide kudos, if my explanation is good enough.
