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Seven family members are seated around their circular dinner

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by BTGmoderatorAT » Sun Nov 26, 2017 3:59 am
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

I'm confused how to set up the formulas here. Can any experts help?
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Source: — Problem Solving |

by GMATGuruNY » Sun Nov 26, 2017 4:08 am
For circular arrangements:
1. Place someone at the table.
2. Count the number of ways to arrange the REMAINING people.
ardz24 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
Once Michael has been placed at the table:
Number of options for Bobby = 2. (To the left or right of Michael.)
Number of ways to arrange the remaining 5 people = 5!.
To combine these options, we multiply:
2 * 5! = 2 * 120 = 240.

The correct answer is B.
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by Scott@TargetTestPrep » Mon Oct 07, 2019 10:21 am
BTGmoderatorAT 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

I'm confused how to set up the formulas here. Can any experts help?
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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