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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.

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 _________________

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...

gmatclubot

Re: PS - combinations
[#permalink]
28 Jun 2006, 19:06

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...