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# combinations

Author Message
Director
Joined: 17 Oct 2005
Posts: 924

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25 Apr 2006, 10:29
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

just a refresher,

how many ways could you arrange the following letters? how do you approach these types agian? thanks
1)AABB

2)ABCDEF

3)AABBCC

4)AAB

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Manager
Joined: 13 Dec 2005
Posts: 224

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Location: Milwaukee,WI

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25 Apr 2006, 10:42
joemama142000 wrote:
just a refresher,

how many ways could you arrange the following letters? how do you approach these types agian? thanks
1)AABB 4!/(2! *2!)

2)ABCDEF 6!

3)AABBCC 6! /(2! *2! *2!)

4)AAB3!/2!

to solve these kinds of problem first count the total no of alphabets say N and in the numerator put N! and if some of them are repeated r1 times , and some other r2 times ........divide the numerator by those factorial . so it becomes N!/(r1! *r2!)

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Manager
Joined: 13 Dec 2005
Posts: 224

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Location: Milwaukee,WI

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25 Apr 2006, 11:07
joemama142000 wrote:
thanks ipc

you are always welcome .

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Senior Manager
Joined: 24 Jan 2006
Posts: 251

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25 Apr 2006, 12:32
a real refresher .. indeed!

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Director
Joined: 04 Jan 2006
Posts: 922

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25 Apr 2006, 18:44
thanks a lot!

that was a good refresher!

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GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5039

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Location: Singapore

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25 Apr 2006, 18:56
1)AABB
Number of ways to arrange any 4 letters = 4!
But A and B are repeated, so we need to consider that there are going to be duplicates. The total number of ways is therefore 4!/2!2! = 6 ways

2)ABCDEF
The number of ways to arrange this is 6! simply because each letter is a different entity.

3)AABBCC
The way to approach this is the same as explained in 1. It's 6!/2!2!2! = 90 ways.

4)AAB
Same as for 3. Number of ways = 3!/2! = 3

Kudos [?]: 425 [0], given: 0

25 Apr 2006, 18:56
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