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How many different four letter words can be formed (the

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How many different four letter words can be formed (the [#permalink] New post 20 Sep 2005, 22:40
How many different four letter words can be formed (the words need not be meaningful) using the letters of the word MEDITERRANEAN such that the first letter is E and the last letter is R?

A. 59
B. 11!/2!*2!*2!
C. 56
D. 23
E. 11!/3!*2!*2!*2!

Please simplify

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 [#permalink] New post 21 Sep 2005, 03:11
I'd go for B

you have 13 letters, two of them are alreadty used, so you have 11! possibilities.

There are two As Ns and Es so you have to divide by 2!*2!*2!
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 [#permalink] New post 21 Sep 2005, 08:35
I came up with C as an answe. Explaination as follows

1) Ignoring repetitions in all there 8 letters in total.
2) First and last letters are E & R
3) We need to find number of ways middle 2 letter combination can be generated.

As it is not mentioned that E & R can not appear in these two letters I calculated 8P2 (permutation) for an answer.

Please let me know if I am wrong.
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 [#permalink] New post 21 Sep 2005, 09:05
MBAHopeful wrote:
I came up with C as an answe. Explaination as follows

1) Ignoring repetitions in all there 8 letters in total.
2) First and last letters are E & R
3) We need to find number of ways middle 2 letter combination can be generated.

As it is not mentioned that E & R can not appear in these two letters I calculated 8P2 (permutation) for an answer.

Please let me know if I am wrong.


why are you taking 8 letters only. Isnt 11 letters left to be chosen to fit 2 middle numbers?

the numbers left for middle spots:
E- 2
A - 2
N - 2
R - 1
4 - single occurance letters

we need to find all distinct 2 letter combinations among them. Not sure how to calculated it. Initially i thought it would be:
11p2/2! * 2! * 2!

but calculation results in fraction.
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 [#permalink] New post 21 Sep 2005, 10:26
duttsit wrote:
MBAHopeful wrote:
I came up with C as an answe. Explaination as follows

1) Ignoring repetitions in all there 8 letters in total.
2) First and last letters are E & R
3) We need to find number of ways middle 2 letter combination can be generated.

As it is not mentioned that E & R can not appear in these two letters I calculated 8P2 (permutation) for an answer.

Please let me know if I am wrong.


why are you taking 8 letters only. Isnt 11 letters left to be chosen to fit 2 middle numbers?

the numbers left for middle spots:
E- 2
A - 2
N - 2
R - 1
4 - single occurance letters

we need to find all distinct 2 letter combinations among them. Not sure how to calculated it. Initially i thought it would be:
11p2/2! * 2! * 2!

but calculation results in fraction.


I thought we need to ignore multiple occurrences of the same letter in the word and hence I came up with number 8. Let me know if my thinking is wrong.
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Re: P&C [#permalink] New post 21 Sep 2005, 12:08
I get 59

one E and R are fixed besides that there are 11 letters for two places out of that 8 are distinct so they can be used to produce

8*7 = 56 combinations

plus three letters are repeated twice so they can make 3 more words..

so the answer is 56+3 = 59
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 [#permalink] New post 23 Sep 2005, 00:09
I agree with ranga41, I read the question wrong and thought they were asking for 8 letter words.

Answer is A
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 [#permalink] New post 02 Jan 2006, 10:44
I got 59 too. Except I used a different approact.

We have 3 letter with 2 occurence and 5 with single. To pick 2 letters, I can do it in two different ways.

1. pick 3 repeated then any 8.
2. pick 5 single and then any 7.

result 3*8 + 5*7 = 59.

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Re: P&C [#permalink] New post 22 Jan 2006, 09:53
ranga41 wrote:
I get 59

one E and R are fixed besides that there are 11 letters for two places out of that 8 are distinct so they can be used to produce

8*7 = 56 combinations

plus three letters are repeated twice so they can make 3 more words..

so the answer is 56+3 = 59


Can anyone please explain why we are adding 3 here?

The word formed by two 'different E's' will read the same and should be considered as a same word.


Also,

out of 13, 2 are used up E and R

you are left with 11 alphabets with two N's, 2 A's and 2 E's to be filled in 2 places

should it not be

11x10 -3 = 107
words
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 [#permalink] New post 22 Jan 2006, 12:02
old_dream,

Obviously I too didn't get this right the first time.

But here is what I think:

MEDITERRANEAN has

13 letters with
3 - E's
2 - R's
2 - N's
2 - A's

The words to be formed are of the type E _, _, R

Consider all the single characters withour repititions i.e, we need to choose from : (M, D, I, T, R) + 1 from each of (E, N, A) = 8 characters.

So No of words = 8*7 = 56

Now we had excluded the possibility of (2 Es, 2Ns, 2A's)

So we have only 3 more words to consider:
EAAR
ENNR
EEER

Hence 56+3 = 59
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  [#permalink] 22 Jan 2006, 12:02
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