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In how many ways can 6 people, A, B, C, D, E, F be seated

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In how many ways can 6 people, A, B, C, D, E, F be seated [#permalink] New post 07 Feb 2012, 20:34
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

48% (02:31) correct 52% (01:29) wrong based on 58 sessions
In how many ways can 6 people, A, B, C, D, E, F be seated in a row such that C and D are not seated next to each other as well as A and B are not seated next to each other?

A 384
B 396
C 576
D 624
E 696
[Reveal] Spoiler: OA
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Kudos [?]: 30960 [1] , given: 3265

Re: Permutations and Combinations [#permalink] New post 08 Feb 2012, 02:44
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Smita04 wrote:
In how many ways can 6 people, A, B, C, D, E, F be seated in a row such that C and D are not seated next to each other as well as A and B are not seated next to each other?

A 384
B 396
C 576
D 624
E 696


Total # of arrangements of 6 people in a row is 6!;

Now, let's find # of arrangements where C and D as well as A and B ARE seated next to each other. Consider them as a single unit: {AB}, {CD}, E, F. Now, these 4 units can be arranged in 4!, but we should multiply this number by 2*2 as A and B, as well as C and D within their unit can be arranged in two ways: {AB} or {BA}, and {CD} or {DC}. Hence total for this case is 4!*2*2;

So, {# of arrangements}={total}-{restriction}=6!-4!*2*2=624.

Answer: D.
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Kudos [?]: 5607 [0], given: 166

Re: Permutations and Combinations [#permalink] New post 07 Feb 2012, 21:08
Expert's post
Smita04 wrote:
In how many ways can 6 people, A, B, C, D, E, F be seated in a row such that C and D are not seated next to each other as well as A and B are not seated next to each other?

A 384
B 396
C 576
D 624
E 696


This is very similar to a question I discussed in a post a week back. Let me give you the link of the post which explains the concept in detail. Perhaps you can check it out and see if you can figure out the answer of this one on your own! I have discussed two methods in it.

http://www.veritasprep.com/blog/2012/01 ... e-couples/
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Joined: 10 Jan 2010
Posts: 193
Location: Germany
Concentration: Strategy, General Management
Schools: IE '15 (M)
GMAT 1: Q V
GPA: 3
WE: Consulting (Telecommunications)
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Kudos [?]: 22 [0], given: 7

Re: In how many ways can 6 people, A, B, C, D, E, F be seated [#permalink] New post 08 Feb 2012, 04:18
Number of total arrangements = 6!
Restriction 1= AB & CD not next to each other --> let say AB and CD are considered as one unit, respectively
Restriction 2= AB is not the same as BA + CD is not the same as DC --> the number will increase by 2*2

Total number of arrangements - Number out of restrictions = Result

6! - (4!*2*2) = 720 - (24*2*2) = 624

Answer D
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Re: In how many ways can 6 people, A, B, C, D, E, F be seated [#permalink] New post 25 May 2013, 03:56
Expert's post
Re: In how many ways can 6 people, A, B, C, D, E, F be seated   [#permalink] 25 May 2013, 03:56
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