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Mediterranean

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by g_beatthegmat » Sun Jun 15, 2008 9:20 pm
How many different four letter words can be made using the letters of the word MEDITERRANEAN, such that the first letter is E and the last letter is R?

(a) 11!/(2!*2!*2!)
(b) 59
(c) 56
(d) 23
(e) 11!/(3!*2!*2!*2!)
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by atlantic » Sun Jun 15, 2008 10:39 pm
There are 13 letters in the word but only 8 are different.

Since the first and the last letter are fixed we are left with only two open cases.

Hence, 8!/6! = 56. Choose C.

OA?
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by g_beatthegmat » Mon Jun 16, 2008 1:32 am
The OA is B
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by atlantic » Mon Jun 16, 2008 3:36 am
Sure? Gosh! How can that be? Such an odd number! lol
I've not a clue how it can be 59 and why my reasoning is incorrect. It seems so simple that I would suggest that the solution is wrong, but I've seen a lot of odd things in GMAT.

Any expert please......

BTW: g_beatthegmat, what is your solution and reasoning (if you have any)?
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Re: Mediterranean

by Ian Stewart » Mon Jun 16, 2008 5:11 am
g_beatthegmat wrote:How many different four letter words can be made using the letters of the word MEDITERRANEAN, such that the first letter is E and the last letter is R?
We know the word looks like: E _ _ R.

We have used an E and R, so we have eight different letters left:
A,A,D,E,E,I,M,N,N,R,T

There are two possibilities:

-The two remaining letters are different from each other (e.g. EADR, EIER, etc). We have 8 choices for the first letter and 7 for the second; 8*7 = 56 different words.

-The two remaining letters are identical (i.e. EAAR, EEER, ENNR). There are only three pairs of identical letters; 3 different words.

56 + 3 = 59.
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by g_beatthegmat » Tue Jun 17, 2008 10:30 pm
thanks Ian! These are the kind of GMAT tricks we need to be careful of.

Thanks!
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Re: Mediterranean

by durgesh79 » Tue Jun 17, 2008 10:38 pm
Ian Stewart wrote:
g_beatthegmat wrote:How many different four letter words can be made using the letters of the word MEDITERRANEAN, such that the first letter is E and the last letter is R?
We know the word looks like: E _ _ R.

We have used an E and R, so we have eight different letters left:
A,A,D,E,E,I,M,N,N,R,T

There are two possibilities:

-The two remaining letters are different from each other (e.g. EADR, EIER, etc). We have 8 choices for the first letter and 7 for the second; 8*7 = 56 different words.

-The two remaining letters are identical (i.e. EAAR, EEER, ENNR). There are only three pairs of identical letters; 3 different words.

56 + 3 = 59.
Ian, you rock !!!! 8)
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