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

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Difficulty:

65% (hard)

Question Stats:

48% (02:27) correct
52% (01:44) wrong based on 138 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?

Re: Permutations and Combinations [#permalink]
07 Feb 2012, 21:08

Expert's post

1

This post was BOOKMARKED

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.

Re: Permutations and Combinations [#permalink]
08 Feb 2012, 02:44

Expert's post

5

This post was BOOKMARKED

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.

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

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

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?

gmatclubot

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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