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In how many ways can the letters of the word EDUCATION be

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In how many ways can the letters of the word EDUCATION be [#permalink] New post 13 Mar 2005, 18:14
In how many ways can the letters of the word EDUCATION be rearranged so that the relative position of the vowels and consonants remain the same as in the word EDUCATION?

(a) 9!/4
(b) 9!/(4!*5!)
(c) 4!*5!
(d) 9!/2!
(e) None of these
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 [#permalink] New post 13 Mar 2005, 19:31
Vowels: E U A I O
Consonants D C T N

We have 9 spaces, space 1,3,5,7,8 must be occupied by vowels
space 2,4,6,9 must be occupied by consonants

So total number of ways = 9!/5!4! --> (B)
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 [#permalink] New post 14 Mar 2005, 13:04
Explanation please. i got that answer also 9!/4!*5!

but i am not sure if i used the right method.
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 [#permalink] New post 15 Mar 2005, 14:20
9!/(5!*4!)
9! # of ways 9 have to be positioned in 9 spaces
5!# of ways 5 have to be positioned in 5 spaces
4!#of ways 4 have to be positioned in 4 spaces
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 [#permalink] New post 15 Mar 2005, 21:04
You're all ready to take the test ! OA is 9!/(5!*4!)
:done
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Please explain [#permalink] New post 16 Mar 2005, 02:17
Rupstar wrote:
9!/(5!*4!)
9! # of ways 9 have to be positioned in 9 spaces
5!# of ways 5 have to be positioned in 5 spaces
4!#of ways 4 have to be positioned in 4 spaces



Alright, this must be really dumb of me, but I still don't get this.

We have 9 positions, of which some specific 5 positions are for vowels and 4 for consonents.

So far so good.

5 distinct vowels can be placed in 5 places in 5! ways.
4 distinct consonents can be placed in 4 places in 4! ways.

Total combinations of consonents and vowels = 5! * 4!

When the consonents and vowels can not be taken as one entity for the purpose of re-arranging, why would the term 9! enter the answer, and why would 4! * 5! come in the denominator?

The answer 9!/(4!*5!) is inverse of the probability of finding the consonents and vowels in the same place as EDUCATION when its alphabets are randomly rearranged.
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 [#permalink] New post 06 Apr 2005, 10:31
Agree with Kapslock that the answer given is the inverse of the probability.
Why is the answer 9!/(5!*4!)?.
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 [#permalink] New post 06 Apr 2005, 19:09
No. 9! is the total # of ways to arrange 9 letters. But we're told 5 of these letters (vowels) have to be arranged in a certain manner, and 4 of these leters (cosonants) have to be arranged in a certain manner.
So 9!/5!4! is the # of ways you can re-arrange the 9 letters given the conditions above.
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 [#permalink] New post 06 Apr 2005, 19:25
I don't get this in fact it's like we are doing a 9C5 with this result. This answer is confusing and i see Kapslock's point
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 [#permalink] New post 07 Apr 2005, 09:00
Hmmm it would be 4!*5! for me.

If it was asking about probability, then it would be 4!*5!/9!
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 [#permalink] New post 07 Apr 2005, 10:44
HongHu wrote:
Hmmm it would be 4!*5! for me.

If it was asking about probability, then it would be 4!*5!/9!


I have re-read the question like 5 times to ensure if it asks for probability and it doesn't
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 [#permalink] New post 07 Apr 2005, 11:26
I know. And even if it were it still isn't 9!/4!*5!.
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 [#permalink] New post 07 Apr 2005, 11:59
HongHu wrote:
I know. And even if it were it still isn't 9!/4!*5!.


I agree with Hong. Since we can not ignore the condition "the letters of the word EDUCATION be rearranged so that the relative position of the vowels and consonants remain the same"

Since 9! will not take care of the relative position of the vowels and consonents. i.e., 9! is just all possible ways we can arrange the alphabets in "EDUCATION".

9!/(4!*5!) will still not take into account the relative position. If I am not wrong we divide for repeating vowels/consonants (I am not sure if that concept is relevent for this problem).

<B> E </B>D<B> U </B>C<B> A </B>T<B> I O </B>N

since we have 5 places for Vowels => we can arrange 5 vowels in 5! ways.
Since we have 4 places for consonents => we can arrange 4 consonents in 4! ways...

C) 4!*5! may be the correct answer.

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 [#permalink] New post 07 Apr 2005, 12:37
Oh I got it now. It was saying that the order of the vowels and consonants should not be changed. Therefore we have 9! total outcomes, but 4! ways to change the order of the consonants and 5! ways to change the order of vowels so you need to divide 9! by 4!*5!.

The correct answer is indeed 9!/4!*5!.
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 [#permalink] New post 08 Apr 2005, 01:31
HongHu wrote:
Oh I got it now. It was saying that the order of the vowels and consonants should not be changed. Therefore we have 9! total outcomes, but 4! ways to change the order of the consonants and 5! ways to change the order of vowels so you need to divide 9! by 4!*5!.

The correct answer is indeed 9!/4!*5!.



Hmm correct me if I am wrong, but 'I think I can understand the question and the OA, if they're not talking about the relative position of vowels and consonants but the relative ordering of the vowels and consonants?

Which means we don't have to fix the positions of vowels in EDUCATION (like position 1, 3, 5, 7 and 8 for vowels and 2, 4, 6 and 9 for consonants), but just ensure that the order of E, U, A, I, O is preserved in each order, as is the order of D, C, T and N.

Right said?
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 [#permalink] New post 08 Apr 2005, 07:55
You are exactly right! The question is not worded well. However I'm glad that we've figure out what it is about. :-D
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  [#permalink] 08 Apr 2005, 07:55
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