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In how many ways can the letters of the word EDUCATION be [#permalink]
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

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

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. _________________

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.

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...

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!. _________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

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.