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How many different four-letter words can be formed (the words don't [#permalink]

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12 Dec 2007, 11:15

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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!)

Re: How many different four-letter words can be formed (the words don't [#permalink]

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13 Dec 2007, 02:41

young_gun wrote:

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, could you explain that to me so that I can easily understand?? I am very bad at perms!

I understand point 1 and point 2 as well...but why N=(3*8+5*7)?thanks

for letters of E,A,N at second position we have 8 cases for third one. So, 3*8
for letters of M,D,I,T,R at second position we have 7 cases for third one (we cannot use, for example, M twice). So, 5*7

Re: How many different four-letter words can be formed (the words don't [#permalink]

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13 Dec 2007, 13:32

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young_gun wrote:

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!

We have 11 letters after E and R occupied their places. But E, A and N show up twice each. So we have 8 distinct letters for 2 places.
For the second place - 8 letters
for the third - 7 letters
Number of variants - 8*7=56, but we have to take into account additional 3 variants with double letters EAAR, ENNR, EEER.
So the ultimate calculation is 56+3=59

Hi Bunuel, could you please explain the solution in your words ?

Thanks.

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!)

E - - R

We are left with the following 11 letters: {M, D, I, T, R, EE, AA, NN} out of which 8 are distinct: {M, D, I, T, R, E, A, N}.

We should consider two cases: 1. If the two middle letters are the same, we'd have 3 words: EEER, EAAR and ENNR.

2. If the two middle letters are distinct, then we are basically choosing 2 letters out of 8 when the order of the selection matters, so it's 8P2 = 56.

In the above problem, if the letters of the word MEDITERRANEAN are allowed to be used multiple times irrespective of their count in the parent word (commonly referred as ‘repetition’ in the P&C parlance), the answer would change. Let me explain the solution for such a case.

We need to fill the 2nd and the 3rd place with letters present in the word MEDITERRANEAN. Since, there are 8 different letters (M, E, D, I, T, R, A, N) in the word MEDITERRANEAN, the 2nd place can be filled with 8 possible letters and the 3rd place can also be filled with 8 possible letters (because, in the case we are discussing here, the letters can be used multiple times, even if they are present only once in the word MEDITERRANEAN).

So, we will have a total of 8*8 = 8^2= 64 possible set of words

Similarly, if the above case is extended to the first and the last letter as well (i.e. we don’t have the constraint of having ‘E’ as the first letter and ‘R’ as the last letter), we will have 8^4 possible sets of words which we can form from the word MEDITERRANEAN.

The key here is to be careful on two points:

•Whether letters can be used more than their count in the parent word, in this case MEDITERRANEAN.

•If yes, then we need to focus only on different letters present in the parent word, in this case the 8 different letters in the parent word MEDITERRANEAN.

Re: How many different four-letter words can be formed (the words don't [#permalink]

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11 Apr 2015, 03:04

I used a diffrent method:

we can also solve this question with combinatorics fairly easy:

after E and R are set as the first and the last letters we are left with the two middle ones.

since both E and R show up more then once we can still use all the original letters for the two remeaining blanks.

actually our bank of letters will now look as so: M=1 E=2 D=1 I=1 T=1 R=1 A=2 N=2

if all remaining letters would have shown up just once the answer would have been: #=8P2=8!/(8-2)!=56

but since we are left with 3 letters that show up more then once (E,A,N) we need to add the possibilty of using the same letter twice, meaning: #=8P2+3=59

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Re: How many different four-letter words can be formed (the words don't [#permalink]

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04 Jun 2017, 18:14

Hello all, First, sorry for posting on this post again, idk if it is allowed I understood the correct answer, but I cannot figure out why my approach is incorrect. We know the total of letters as well as the total of repeated letters. M - 1 E - 3 D - 1 I - 1 T - 1 R - 2 A - 2 N - 2 And we know that E and R have been already used once, so we now have 2 E's and 1 R. If the question was: how many different 13-letter words can be formed, I would calculate like this: E x x x x x x x x x x x R - > 11! / (2! * 2! * 2!) (and there is an answer for that) As the question if for 4-letter word, and we have just to spaces left, I would just do this: 11 * 10 / (2! * 2! * 2!) I know that this is incorrect (also because it isn't an integer number ). But I do not know why! Thanks!

Re: How many different four-letter words can be formed (the words don't [#permalink]

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05 Jun 2017, 06:53

young_gun wrote:

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!)

We have to foind total ways of filling E _ _ R After taking one E and one R out we get 8 unique letters M E D I T R A N and 3 double letters E A N Taking 2 letters from set 1 and arranging them = \(_{8}C_{2} * 2!\) = 56 Taking 2 letters from set 2 and arranging them(arrangement in this case does not matter) = 1+1+1=3