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

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Senior Manager
Joined: 22 Feb 2018
Posts: 312
At a birthday party, 10 students are to be seated around a circular  [#permalink]

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19 Nov 2019, 03:30
1
4
00:00

Difficulty:

45% (medium)

Question Stats:

48% (01:10) correct 52% (01:28) wrong based on 66 sessions

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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
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Joined: 16 Feb 2015
Posts: 124
Location: United States
Concentration: Finance, Operations
At a birthday party, 10 students are to be seated around a circular  [#permalink]

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Updated on: 19 Nov 2019, 04:04
1
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.
Bunuel bb

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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Posts: 59568
Re: At a birthday party, 10 students are to be seated around a circular  [#permalink]

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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 = $$\frac{favorable}{total}=\frac{9!-8!*2}{9!}=1-\frac{2}{9}=\frac{7}{9}$$

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

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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.

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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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