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

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Manager
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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16 Oct 2008, 06:11
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

SVP
Joined: 29 Aug 2007
Posts: 2472

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

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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.
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

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

Manager
Joined: 12 Feb 2008
Posts: 177

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

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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
Joined: 29 Aug 2007
Posts: 2472

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

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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.
_________________

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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