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

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Manager
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Have 6 words, A, B, C, D, E, F (for example). How many [#permalink] New post 28 Jun 2006, 16:30
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A
B
C
D
E

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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
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 [#permalink] New post 28 Jun 2006, 17:43
number of total possibility = 6! = 6*5*4*3*2*1 = 720

ABDCEF
ABDCFE
ABECDF
ABECFD
ABFCDE
ABFCED

720-6 = 714...
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Re: PS - combinations [#permalink] New post 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
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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Re: PS - combinations [#permalink] New post 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... :oops:
Re: PS - combinations   [#permalink] 28 Jun 2006, 19:06
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Have 6 words, A, B, C, D, E, F (for example). How many

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