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# Another Probability question..

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Manager
Joined: 22 Feb 2009
Posts: 136

Kudos [?]: 149 [0], given: 10

Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)

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21 May 2009, 10:21
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Q. On how many ways can the letters of the word "COMPUTER" be arranged?
Vowels occupy the even positions.

Do we have to select 4 even positions out of 8 position first for solving this question?

Kudos [?]: 149 [0], given: 10

Manager
Joined: 11 Apr 2009
Posts: 160

Kudos [?]: 114 [0], given: 5

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21 May 2009, 13:37
There are 3 vowels and 5 consonants.

Now these 3 vowels occupy positions 2,4,6,8 (4 positions)
so number of ways these 3 vowels can be arranged is 4P3=4!/(4-3)! = 4! = 4*3*2*1=24

Now 5 positions are for the 5 consonants and so can be arranged in 5P5 = 5! = 120

Therefore total number of ways the letters with given constraint can be arranged is : 24*120.

What is the OA?

Kudos [?]: 114 [0], given: 5

Senior Manager
Joined: 08 Jan 2009
Posts: 325

Kudos [?]: 176 [0], given: 5

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22 May 2009, 01:36
Agree with gmatprep09.

4C3 * 3! * 5! = 24 * 120

Kudos [?]: 176 [0], given: 5

Manager
Joined: 22 Feb 2009
Posts: 136

Kudos [?]: 149 [0], given: 10

Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)

### Show Tags

24 May 2009, 04:21
gmatprep09 wrote:
There are 3 vowels and 5 consonants.

Now these 3 vowels occupy positions 2,4,6,8 (4 positions)
so number of ways these 3 vowels can be arranged is 4P3=4!/(4-3)! = 4! = 4*3*2*1=24

Now 5 positions are for the 5 consonants and so can be arranged in 5P5 = 5! = 120

Therefore total number of ways the letters with given constraint can be arranged is : 24*120.

What is the OA?

Good explanation... the OA is same 4*720

Kudos [?]: 149 [0], given: 10

Re: Another Probability question..   [#permalink] 24 May 2009, 04:21
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