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# In how many ways can the letters of the word ARRANGE be

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In how many ways can the letters of the word ARRANGE be [#permalink]

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12 Apr 2006, 19:12
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In how many ways can the letters of the word ARRANGE be rearranged so that neither the two Rs nor the two As can come together?
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12 Apr 2006, 20:48
chillpill wrote:
In how many ways can the letters of the word ARRANGE be rearranged so that neither the two Rs nor the two As can come together?

I'm guessing is: 7! - 2*6! - 5! = 3480

7! number of ways to arrange a word with the letters A-R-R-A-N-G-E

let's make AA = A and
RR = R

then we can have A-R-R-N-G-E
or A-R-A-N-G-E

in 6! ways

and A-R-N-G-E in 5! ways
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12 Apr 2006, 20:54
conocieur wrote:
chillpill wrote:
In how many ways can the letters of the word ARRANGE be rearranged so that neither the two Rs nor the two As can come together?

I'm guessing is: 7! - 2*6! - 5! = 3480

7! number of ways to arrange a word with the letters A-R-R-A-N-G-E

let's make AA = A and
RR = R

then we can have A-R-R-N-G-E
or A-R-A-N-G-E

in 6! ways

and A-R-N-G-E in 5! ways

A-R-R-A-N-G-E

i think we can AA togather in 2 (6) x 5! ways = 2x6! ways.
similarly RR can also be done in the same way but i guess RR already comes in AA's arrangements.... so
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12 Apr 2006, 21:17
Here is my shot at it...

Total - {2Rs together + 2As together - RRAA all of them together}
.. beacuse both the cases where 2Rs/2As are together also include RRAA. We need to subtract those... Sort of Venn diagram problem
= 7!/(2!*2!) - {(2*6!/2!) - 5! }
= 660

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12 Apr 2006, 21:24
giddi77 wrote:
Here is my shot at it...

Total - {2Rs together + 2As together - RRAA all of them together}
.. beacuse both the cases where 2Rs/2As are together also include RRAA. We need to subtract those... Sort of Venn diagram problem
= 7!/(2!*2!) - {(2*6!/2!) - 5! }
= 660

but remember AA ( or RR) can placed in two ways.. for AA, lets say Aa; we can put Aa two ways. so you are missing 2.

confusing.....
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13 Apr 2006, 00:16
giddi77 wrote:
Here is my shot at it...

Total - {2Rs together + 2As together - RRAA all of them together}
.. beacuse both the cases where 2Rs/2As are together also include RRAA. We need to subtract those... Sort of Venn diagram problem
= 7!/(2!*2!) - {(2*6!/2!) - 5! }
= 660

yeah yeah yeah this is correct. thanks
13 Apr 2006, 00:16
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