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Permutation : word DELETED

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Senior Manager
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Permutation : word DELETED [#permalink]

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New post 20 Mar 2009, 16:09
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Let each different arrangement of all the letters of DELETED be called a word. In how many of these words will the D's be separated?

I can get the number by first finding the total number of words which is 7!/(3!*2!) = 420 and then subtracting from it the total number of words that has D's together (this number is 6!/3! = 120). Can someone explain how I can find the result through regular counting process?

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Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014
Re: Permutation : word DELETED [#permalink]

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New post 27 Mar 2009, 11:24
420 is correct.
now if we take DD as one entity then we have 6! ways, but within this D is repeating so we get 6!/2! ...why do we take 6!/3! . Please explain.
sanjay_gmat wrote:
Let each different arrangement of all the letters of DELETED be called a word. In how many of these words will the D's be separated?

I can get the number by first finding the total number of words which is 7!/(3!*2!) = 420 and then subtracting from it the total number of words that has D's together (this number is 6!/3! = 120). Can someone explain how I can find the result through regular counting process?

Kudos [?]: 843 [0], given: 18

Senior Manager
Senior Manager
avatar
Joined: 06 Jul 2007
Posts: 275

Kudos [?]: 53 [0], given: 0

Re: Permutation : word DELETED [#permalink]

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New post 27 Mar 2009, 14:03
Economist wrote:
420 is correct.
now if we take DD as one entity then we have 6! ways, but within this D is repeating so we get 6!/2! ...why do we take 6!/3! . Please explain.
sanjay_gmat wrote:
Let each different arrangement of all the letters of DELETED be called a word. In how many of these words will the D's be separated?

I can get the number by first finding the total number of words which is 7!/(3!*2!) = 420 and then subtracting from it the total number of words that has D's together (this number is 6!/3! = 120). Can someone explain how I can find the result through regular counting process?



Well, the six elements in second case (both D's together) are :

1 - DD
2 - E
3 - E
4 - E
5 - L
6 - T

in this group, there are 3 repetitive elements, hence total number of permutations = 6!/3!.

I don't think you need to divide by 2! because the 2Ds are considered one single group.

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Manager
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Re: Permutation : word DELETED [#permalink]

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New post 27 Mar 2009, 21:13
sanjay_gmat wrote:
Let each different arrangement of all the letters of DELETED be called a word. In how many of these words will the D's be separated?

I can get the number by first finding the total number of words which is 7!/(3!*2!) = 420 and then subtracting from it the total number of words that has D's together (this number is 6!/3! = 120). Can someone explain how I can find the result through regular counting process?


I'm probably missing something, but could you please confirm you answer is 120?
Because I got 60 as the final answer.

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Manager
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Re: Permutation : word DELETED [#permalink]

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New post 06 Apr 2009, 23:23
So..Sanjay..Is the answer 420-120 = 300 ways that the D's will be separated?

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Re: Permutation : word DELETED   [#permalink] 06 Apr 2009, 23:23
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