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On six seating chairs including one red chair around a big r

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by Max@Math Revolution » Tue Oct 04, 2016 6:12 am
On six seating chairs including one red chair around a big round table , A, B, C, D, E, F each sit on a chair. A must sit on the red chair and B must be beside A. How many seating arrangement are?
A. 36 B. 40 C. 44 D. 48 E. 64
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by GMATinsight » Fri Oct 14, 2016 10:54 pm
Max@Math Revolution wrote:On six seating chairs including one red chair around a big round table , A, B, C, D, E, F each sit on a chair. A must sit on the red chair and B must be beside A. How many seating arrangement are?
A. 36 B. 40 C. 44 D. 48 E. 64
A has only one place to sit (Red chair) = 1 way of sitting (Fixed chair)

B has two options to sit next to A = 2 ways

Temaining 4 have 4! options to sit on 4 vacant chairs = 4! ways

Total options of all of them to sit = 1*2*4! = 48

Answer: option D
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by fiza gupta » Sat Oct 15, 2016 3:46 am
Straight line =>total arrangements = n!
Round Table ->total arrangements = (n-1)!
but A and B have to sit together , let AB be one pair
AB,C,D,E,F = 5
= (5-1)! = 4! = 24
AB can arrange themselves in 2 ways
= 24*2 = 48 ways
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by Brent@GMATPrepNow » Sat Oct 15, 2016 7:14 am
Max@Math Revolution wrote:On six seating chairs including one red chair around a big round table , A, B, C, D, E, F each sit on a chair. A must sit on the red chair and B must be beside A. How many seating arrangement are?

A. 36
B. 40
C. 44
D. 48
E. 64
Take the task of seating the 6 people and break it into stages.

We'll begin with the most restrictive stage.

Stage 1: Seat person A
Since person A MUST sit in the red chair, we can complete stage 1 in only 1 way

Stage 2: Seat person B
Since person B MUST sit beside person A, we can seat person B to the LEFT of person A or to the RIGHT of person A.
So, we can complete stage 2 in 2 ways

Stage 3: Seat person C
There are now 4 seats remaining, so we can complete stage 3 in 4 ways

Stage 4: Seat person D
There are now 3 seats remaining, so we can complete stage 4 in 3 ways

Stage 5: Seat person E
There are now 2 seats remaining, so we can complete stage 5 in 2 ways

Stage 6: Seat person F
There is now 1 seat remaining, so we can complete stage 6 in 1 way

By the Fundamental Counting Principle (FCP), we can complete all 6 stages (and thus seat all 6 people) in (1)(2)(4)(3)(2)(1) ways ([spoiler]= 48 ways[/spoiler])

Answer: D
--------------------------

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch this video: https://www.gmatprepnow.com/module/gmat- ... /video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/permutation-t122873.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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