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

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Manager
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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
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A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

46% (02:22) correct 54% (01:44) wrong based on 91 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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Re: Permutations and Combinations [#permalink] New post 07 Feb 2012, 21:08
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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


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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Kudos [?]: 40977 [0], given: 5653

Re: Permutations and Combinations [#permalink] New post 08 Feb 2012, 02:44
Expert's post
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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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Manager
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Joined: 10 Jan 2010
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Location: Germany
Concentration: Strategy, General Management
Schools: IE '15 (M)
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Kudos [?]: 23 [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
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Re: In how many ways can 6 people, A, B, C, D, E, F be seated [#permalink] New post 18 Apr 2015, 16:33
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Re: In how many ways can 6 people, A, B, C, D, E, F be seated [#permalink] New post 19 Apr 2015, 02:57
6 people in the row is 6!=720

glue method is best , so consider A&B and C&D sit together

now we have 4!=24, but 2 people in one pair and 2 people in another pair can sit in 4 positions, so 24*4=96

desirable condition is 720-96=624

D
Re: In how many ways can 6 people, A, B, C, D, E, F be seated   [#permalink] 19 Apr 2015, 02:57
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