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Arrange letters of the word

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Arrange letters of the word [#permalink] New post 20 May 2011, 05:55
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

50% (01:34) correct 50% (00:09) wrong based on 10 sessions
In how many ways can the letters of the word JUPITER be arranged in a row so that the vowels appear in alphabetic order?

(A) 736 (B) 768 (C) 792 (D) 840 (E) 876


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Re: Arrange letters of the word [#permalink] New post 20 May 2011, 07:12
three letters can be arranged in 3! ways.
only one combination EIU is required.

7 letters can be arranged in 7! ways.

thus 7!/ 3! * 1 = 840.

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Re: Arrange letters of the word [#permalink] New post 20 May 2011, 07:51
@amit2k9, could you please explain this in more detail ?
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Re: Arrange letters of the word [#permalink] New post 20 May 2011, 10:10
1. Consider EIU as one. then total possible combination of words = 5* 4! = 120

2. Consider EI and IU together total possible combinations = [5 (positions for EI out of 7) * 5! (rest of the letters)] +

[ 5 (positions for IU)* 5! (rest of the letters)]

Now in half of the combinations U will be ahead of EI and E will be ahead of IU.

So 1/2[5 (positions for EI out of 7) * 5! (rest of the letters)] +1/2 [ 5 (positions for IU)* 5! (rest of the letters)]

total will give = 300 + 300 = 600.

3. Consider E-I-U-- with one letter spacing in between = 3(positions of EIU shifting) * 4! (letters to be arranged) = 72

4. Consider E-I--U- and E---I-U with two letter spacing between IU and EI = 2 * 4! = 48

thus totaling 120 + 600 + 72 + 48 = 840.


Alternately , Consider

all possible letter arrangements = 7!

only one combination out of 3! combinations possible for EIU. meaning 3 are taken in a group,but they are not alike.
Means total combinations remain 7! and not 5! ( EIU together + 4 letters).

So, essentially it is 7P3 =
Hence 7!/ 3! = 840
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Re: Arrange letters of the word [#permalink] New post 20 May 2011, 12:51
Amt2k9,

Conceptwise, I did not get your 7!/3!. If the word was, Let's say JUPUTUR, number of ways the letter can be arranged is 7!/3! ways (total ! / identical !). You used the same concept where the letters are not identical. You mentioned that those are not alike, but still not clear on why we still can do 7!/3!.
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Re: Arrange letters of the word [#permalink] New post 20 May 2011, 18:20
Darn ! I am not Amt2k9. I think it is pretty obvious that in 1/6th of the ways out of 7! the vowels will be ordered. This reasoning is parallel to the situation when we have 3 people lets say A B C sitting in a row and we want to know the ways in which they will be alphabetical order. There is just one way ie. 3!/6=1 or 3!/3!

The division by 3! is NOT owned by the concept of weeding out the duplicates. We will apply it when we need it :wink:

bellcurve wrote:
Amt2k9,

Conceptwise, I did not get your 7!/3!. If the word was, Let's say JUPUTUR, number of ways the letter can be arranged is 7!/3! ways (total ! / identical !). You used the same concept where the letters are not identical. You mentioned that those are not alike, but still not clear on why we still can do 7!/3!.
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Re: Arrange letters of the word [#permalink] New post 20 May 2011, 22:27
bellcurve wrote:
Amt2k9,

Conceptwise, I did not get your 7!/3!. If the word was, Let's say JUPUTUR, number of ways the letter can be arranged is 7!/3! ways (total ! / identical !). You used the same concept where the letters are not identical. You mentioned that those are not alike, but still not clear on why we still can do 7!/3!.



Since the order of vowels will always remain the same despite these occupying different positions -> if we assume each vowel as X then our question is same as asking "arrange JPTRXXX" => in all 7!/3! ways
=> Choice (4) is the right answer
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Re: Arrange letters of the word   [#permalink] 20 May 2011, 22:27
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