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# Have 6 words, A, B, C, D, E, F (for example). How many

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Manager
Joined: 04 Dec 2005
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Have 6 words, A, B, C, D, E, F (for example). How many [#permalink]  28 Jun 2006, 16:30
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Have 6 words, A, B, C, D, E, F (for example). How many combinations are possible such that A is not at position 1, B is not at position 2, and C is not position 4, how many possibilities.
Manager
Joined: 19 Apr 2006
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number of total possibility = 6! = 6*5*4*3*2*1 = 720

ABDCEF
ABDCFE
ABECDF
ABECFD
ABFCDE
ABFCED

720-6 = 714...
CEO
Joined: 20 Nov 2005
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Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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Re: PS - combinations [#permalink]  28 Jun 2006, 19:02
withme wrote:
Have 6 words, A, B, C, D, E, F (for example). How many combinations are possible such that A is not at position 1, B is not at position 2, and C is not position 4, how many possibilities.

Total = 6!

When A is at pos1 = 5!
When B is at pos2 = 5!
When C is at pos4 = 5!

Result = 6!-3*5! = 720-360 = 360
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Manager
Joined: 19 Apr 2006
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Re: PS - combinations [#permalink]  28 Jun 2006, 19:06
ps_dahiya wrote:
withme wrote:
Have 6 words, A, B, C, D, E, F (for example). How many combinations are possible such that A is not at position 1, B is not at position 2, and C is not position 4, how many possibilities.

Total = 6!

When A is at pos1 = 5!
When B is at pos2 = 5!
When C is at pos4 = 5!

Result = 6!-3*5! = 720-360 = 360

Thanks, I believe that I miss read the question. I thought the question was asking something else...
Re: PS - combinations   [#permalink] 28 Jun 2006, 19:06
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