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# How many different four-letter words can be formed (the

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VP
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How many different four-letter words can be formed (the [#permalink]  21 Feb 2005, 07:52
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How many different four-letter words can be formed (the words don't need to make sense) 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!/(2!*2!*2!*3!)
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Could someone explain the soln..... Thanks.
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That's right. You have to divide off the repeating elements.

Think of it this way, how many ways are there to arrage 3 alphbets:
ABA

We can have ABA, BAA, and ABA again and finally AAB. But note that two of them are repeated due to the fact we have repeated letters, 'A'.

So the way to do this is to divide off the repeating elements. In my example, this would be 3!/2 = 3. (i.e. ABA, BAA, AAB)
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Re: PS - Combination [#permalink]  23 Feb 2005, 19:56
MEDITERRANEAN

total 13 letters

We want four letter words that the first letter is E and the last letter is R.

So we need to get the two middle letters from the 11 letters (13-E and R): MDITERANEAN
We have 8 different letters with E, A, N repeated twice.

So it would be P(8,2)+3=56+3=59

(A)
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I mean I understand that there are 11! possibilities for two spaces.

There is also 3 letters that are repeated twice (E,A, and N)

Now what I don't understand is where the 3! comes from.

currently I see this equation 11!/(2!*2!*2!)

I realize there is more, but how do we justify the *3! in the denominator?
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Re: PS - Combination [#permalink]  24 Feb 2005, 09:30
HongHu wrote:
MEDITERRANEAN

total 13 letters

We want four letter words that the first letter is E and the last letter is R.

So we need to get the two middle letters from the 11 letters (13-E and R): MDITERANEAN
We have 8 different letters with E, A, N repeated twice.

So it would be P(8,2)+3=56+3=59

(A)

Hong I followed the highlighted part , but then how come +3?????????
Could you plz explain that?

Last edited by Vijo on 25 Feb 2005, 02:54, edited 1 time in total.
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okay [#permalink]  24 Feb 2005, 11:32
But I thought since the word starts with E and ends with R, technically we only have 2 E's that can be used. Can someone please explain?
Senior Manager
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Christoph what's the OA on this one?

P(8,2)+3=56+3=59
HongHo, can you plz explain this a lil bit in details?
Manager
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HongHu seems closest - though I still dont completely understand his solution.

Nice qs bTw
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Ok my thought is this. For the middle two letters, you could have two different ones from the eight different letters, or you could have two same ones from the three repeated letters. So the first part is C(8,2), and the second part is 3 (EE, AA, NN).
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Christoph,
Let's get a resolution to this Q.

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Re: PS - Combination [#permalink]  01 May 2005, 12:16
christoph wrote:
How many different four-letter words can be formed (the words don't need to make sense) 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!/(2!*2!*2!*3!)

how many four letter words if letters can be repeated?
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Re: PS - Combination [#permalink]  02 May 2005, 17:02
christoph wrote:
How many different four-letter words can be formed (the words don't need to make sense) 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!/(2!*2!*2!*3!)

this is really a new concept in counting methods (permutation and combination), if hong's approach is the solution/answer.
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Re: PS - Combination [#permalink]  03 May 2005, 01:07
MA wrote:
christoph wrote:
How many different four-letter words can be formed (the words don't need to make sense) 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!/(2!*2!*2!*3!)

this is really a new concept in counting methods (permutation and combination), if hong's approach is the solution/answer.

why ? we have a four-letter-word that begins with a E and ends with a R => E _ _ R. The two spaces in the middle can be occupied by 8 distinct letters, hence 8P2. but its also possible to have repeated letters such as EE, AA and NN. thats why we add 3. i think its just slightly different to a classical permutation-letter-problem.
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If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

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Re: PS - Combination [#permalink]  10 May 2005, 13:56
HongHu wrote:
MEDITERRANEAN

total 13 letters

We want four letter words that the first letter is E and the last letter is R.

So we need to get the two middle letters from the 11 letters (13-E and R): MDITERANEAN
We have 8 different letters with E, A, N repeated twice.

So it would be P(8,2)+3=56+3=59

(A)

You nailed it.
_________________

Its not the fact its your attitude towards the fact

Re: PS - Combination   [#permalink] 10 May 2005, 13:56
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