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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, 21:34

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53% (01:43) correct
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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?

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.

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.

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

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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

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

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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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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.

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]

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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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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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 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.
_________________

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

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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.

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".
_________________

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

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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.
_________________

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

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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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