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example: There are five council members (ABCDE), A and D

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Manager
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Joined: 01 Jan 2008
Posts: 222

Kudos [?]: 62 [0], given: 2

Schools: Booth, Stern, Haas
example: There are five council members (ABCDE), A and D [#permalink]

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New post 16 Oct 2008, 06:11
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example: There are five council members (ABCDE), A and D cannot sit next to each other. How many different possible seating arrangements are there for the five council members?


Is there any other way to solve it except of next way:
5! - 4!2!=72

Kudos [?]: 62 [0], given: 2

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Joined: 29 Aug 2007
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Kudos [?]: 843 [0], given: 19

Re: Combinatorics-one more time [#permalink]

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New post 16 Oct 2008, 07:33
kazakhb wrote:
example: There are five council members (ABCDE), A and D cannot sit next to each other. How many different possible seating arrangements are there for the five council members?


Is there any other way to solve it except of next way:
5! - 4!2!=72


your's is the most easiest way.
_________________

Verbal: http://gmatclub.com/forum/new-to-the-verbal-forum-please-read-this-first-77546.html
Math: http://gmatclub.com/forum/new-to-the-math-forum-please-read-this-first-77764.html
Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html


GT

Kudos [?]: 843 [0], given: 19

Manager
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Joined: 12 Feb 2008
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Kudos [?]: 46 [0], given: 0

Re: Combinatorics-one more time [#permalink]

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New post 16 Oct 2008, 08:12
kazakhb wrote:
example: There are five council members (ABCDE), A and D cannot sit next to each other. How many different possible seating arrangements are there for the five council members?


Is there any other way to solve it except of next way:
5! - 4!2!=72


can you please explain the second part 4!2!

thanks,

Kudos [?]: 46 [0], given: 0

SVP
SVP
User avatar
Joined: 29 Aug 2007
Posts: 2472

Kudos [?]: 843 [0], given: 19

Re: Combinatorics-one more time [#permalink]

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New post 16 Oct 2008, 09:13
elmagnifico wrote:
kazakhb wrote:
example: There are five council members (ABCDE), A and D cannot sit next to each other. How many different possible seating arrangements are there for the five council members?


Is there any other way to solve it except of next way:
5! - 4!2!=72


can you please explain the second part 4!2!

thanks,


lets suppose A and D as X.
No of ways A and D can be put in X = 2
No of ways X, B, C, and E can be put = 4!
so it is 2 (4!) = 48.
_________________

Verbal: http://gmatclub.com/forum/new-to-the-verbal-forum-please-read-this-first-77546.html
Math: http://gmatclub.com/forum/new-to-the-math-forum-please-read-this-first-77764.html
Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html


GT

Kudos [?]: 843 [0], given: 19

Re: Combinatorics-one more time   [#permalink] 16 Oct 2008, 09:13
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example: There are five council members (ABCDE), A and D

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