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Seven family members are seated around their circular dinner

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Seven family members are seated around their circular dinner  [#permalink]

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New post Updated on: 09 Jul 2013, 07:43
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Seven family members are seated around their circular dinner table. If only arrangements that are considered distinct are those where family members are seated in different locations relative to each other, and Michael and Bobby insist on sitting next to one another, then how many distinct arrangements around the table are possible?

A. 120
B. 240
C. 360
D. 480
E. 720
Spoiler: OA

Originally posted by eladshush on 04 Oct 2010, 05:30.
Last edited by Bunuel on 09 Jul 2013, 07:43, edited 1 time in total.
Edited the OA.
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Re: Arrangements around the table  [#permalink]

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New post 04 Oct 2010, 08:06
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eladshush wrote:
Seven family members are seated around their circular dinner table. If only arrangements that are considered distinct are those where family members are seated in different locations relative to each other, and Michael and Bobby insist on sitting next to one another, then how many distinct arrangements around the table are possible?

A. 120
B. 240
C. 360
D. 480
E. 720


Glue Michael and Bobby so that they create one unit, so we would have total of 6 units: {1}{2}{3}{4}{5}{MB} --> # of different arrangements of \(n\) objects around the table (circular arrangements) is is \((n-1)!\), so our 6 objects can be arranged in \((6-1)!=5!\).

On the other hand Michael and Bobby in 2! ways --> total \(5!*2!=240\).

Answer: B.

Similar question also posted by you: ways-to-sit-around-the-table-102187.html
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Re: Arrangements around the table  [#permalink]

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New post 04 Oct 2010, 08:01
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eladshush wrote:
Seven family members are seated around their circular dinner table. If only arrangements that are considered distinct are those where family members are seated in different locations relative to each other, and Michael and Bobby insist on sitting next to one another, then how many distinct arrangements around the table are possible?

A. 120
B. 240
C. 360
D. 480
E. 720


Are you sure you got the right OA on this one ?

As I see it : 7 members, and Michael & Bobby sit together. So treat those two like a single person. Now you have to arrange 6 people around a table. This may be done in 5! or 120 ways

The number of ways to arrange Michael & Bobby is 2 (MB or BM).

So net there are 240 ways or B

I don't think this is a GMAT level question, may be a tad too hard
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Re: Seven family members are seated around their circular dinner  [#permalink]

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New post 29 Jun 2016, 01:25
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eladshush wrote:
Seven family members are seated around their circular dinner table. If only arrangements that are considered distinct are those where family members are seated in different locations relative to each other, and Michael and Bobby insist on sitting next to one another, then how many distinct arrangements around the table are possible?

A. 120
B. 240
C. 360
D. 480
E. 720


Lets consider Michael and Bobby as one individual and fix their position so that all the members do NOT move together while they remain in same order relatively

Now after fixing Michael and Bobby we have 5 other member left to change their positions among themselves which can change positions in 5! ways

but Michael and Bobby and exchange positions between the two in 2! ways'

Hence, Total ways of different arrangements = 5!*2! = 120*2 = 240

Answer: Option B
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Re: Seven family members are seated around their circular dinner  [#permalink]

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New post 09 Oct 2019, 19:04
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eladshush wrote:
Seven family members are seated around their circular dinner table. If only arrangements that are considered distinct are those where family members are seated in different locations relative to each other, and Michael and Bobby insist on sitting next to one another, then how many distinct arrangements around the table are possible?

A. 120
B. 240
C. 360
D. 480
E. 720


If Bobby and Michael must sit next to each other, we treat them as a single entity, and that leaves us with 6 total spots to arrange. Using the circular permutations formula (n - 1)!, we have

(6 - 1)! = 5! = 120 ways to arrange the family members with Bobby and Michael together.

However, we also must include the number of ways to arrange Bobby and Michael, which is 2P2 = 2! = 2.

So, in total, we have:

(6 - 1)! * 2! = 120 x 2 = 240

Answer: B
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Re: Seven family members are seated around their circular dinner   [#permalink] 09 Oct 2019, 19:04
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