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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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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]

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New post 07 Feb 2012, 22:08
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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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Re: Permutations and Combinations [#permalink]

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New post 08 Feb 2012, 03:44
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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Re: In how many ways can 6 people, A, B, C, D, E, F be seated [#permalink]

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New post 08 Feb 2012, 05:18
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]

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New post 18 Apr 2015, 17:33
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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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New post 19 Apr 2015, 03:57
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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New post 16 Oct 2015, 00:22
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?

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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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New post 26 Mar 2017, 22:34
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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New post 15 Jun 2017, 02:30
Bunuel 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


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.

Hi Bunuel,
In my opinion, this question translates into "neither A and B, nor C and D should sit together". I think that the correct answer is not mentioned in the options.
There are four cases possible

1. AB sit together and CD sit together
2. AB sit together but CD don't sit together
3. CD sit together but AB don't sit together
4. neither AB nor CD sit together.

We have to subtract 1,2, and 3 from total possible cases.

I have found the answer to be 336 in this manner.
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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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New post 12 Jul 2017, 01:11
ShashankDave wrote:
Bunuel 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


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.

Hi Bunuel,
In my opinion, this question translates into "neither A and B, nor C and D should sit together". I think that the correct answer is not mentioned in the options.
There are four cases possible

1. AB sit together and CD sit together
2. AB sit together but CD don't sit together
3. CD sit together but AB don't sit together
4. neither AB nor CD sit together.

We have to subtract 1,2, and 3 from total possible cases.

I have found the answer to be 336 in this manner.


Agreed ! The answer choices are incorrect. It should be 336.

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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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New post 12 Jul 2017, 01:18
ShashankDave wrote:
Bunuel 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


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.

Hi Bunuel,
In my opinion, this question translates into "neither A and B, nor C and D should sit together". I think that the correct answer is not mentioned in the options.
There are four cases possible

1. AB sit together and CD sit together
2. AB sit together but CD don't sit together
3. CD sit together but AB don't sit together
4. neither AB nor CD sit together.

We have to subtract 1,2, and 3 from total possible cases.

I have found the answer to be 336 in this manner.


The language might be misleading here but still I translated "as well as" as "and" not as "or".
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

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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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New post 18 Jul 2017, 06:53
Bunuel 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


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.



What if AB are together, but CD are separate. Similarly, if CD are together and AB are separate.
These two cases shouls aldo be taken into account.
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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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New post 18 Jul 2017, 06:57
Hi Bunuel,
In my opinion, this question translates into "neither A and B, nor C and D should sit together". I think that the correct answer is not mentioned in the options.
There are four cases possible

1. AB sit together and CD sit together
2. AB sit together but CD don't sit together
3. CD sit together but AB don't sit together
4. neither AB nor CD sit together.

We have to subtract 1,2, and 3 from total possible cases.

I have found the answer to be 336 in this manner.[/quote]

The language might be misleading here but still I translated "as well as" as "and" not as "or".[/quote]

That's not correct. I have worked on similar problems and I agree what ShashankDave has to say. I am 100% sure about it. I ave spoken to various maths experts, and they've told me the same thing!
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Re: In how many ways can 6 people, A, B, C, D, E, F be seated   [#permalink] 18 Jul 2017, 06:57
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