In how many ways can 6 people, A, B, C, D, E, F be seated : GMAT Problem Solving (PS)
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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]

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07 Feb 2012, 20:34
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51% (02:41) correct 49% (01:45) wrong based on 176 sessions

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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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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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08 Feb 2012, 02:44
Expert's post
5
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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.

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

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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

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

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25 May 2013, 03:56
Bumping for review and further discussion.
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Re: In how many ways can 6 people, A, B, C, D, E, F be seated [#permalink]

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18 Apr 2015, 16:33
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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

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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

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

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15 Oct 2015, 23:22
VeritasPrepKarishma wrote:
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/

@Veritasprep

What is the difference between this question and the one mentioned in the link provided? They seem to be the same to me so why the difference in the approach and consequently, different answers?
Re: In how many ways can 6 people, A, B, C, D, E, F be seated   [#permalink] 15 Oct 2015, 23:22
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