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.

Total = 56 + 3 = 59.

Answer: A.

Hope it's clear.
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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

Please, could you explain that to me so that I can easily understand?? I am very bad at perms![/quote]

we should complete word E _ _ R using set {M-1, E-2 (one E we use as the first letter), D-1, I-1,T-1,R-1 (one R we use as the last letter) ,A-2,N-2}

So, the set consist of 5 single letters and 3 pairs of letters.

1. for second position we have 8 cases (or 5+3)

2. for third position we have either 8 cases (second letter is from a pair) or 7 cases (second letter is single letter).

Therefore,
N=(3*8+5*7)=59

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

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.

Hope it helps!

Regards
Harsh
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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

so the answer is A.

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!

Answer": Option A

Find the solution as attached.

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This is the official answer - however can anyone explain why it can't be done with the permutation formula 11P2 and the dividing for repetition so choice B?

Step 1 of solving this GMAT Permutation Question: Select 2 letters and rearrange them
MEDITERRANEAN is a 13-letter word.
We have to form a 4-letter words that start with 'E' and ends with 'R'.
Therefore, in addition to E and R, we have to find two more letters from the remaining 11 letters.
In these 11 letters, there are 2 Ns, 2Es, and 2As and one each of the remaining 5 letters viz., M, D, I, T, and R.

Step 2 of solving this GMAT Permutation Question: List down the different posssibilities
Of the 11 letters, there are 2 Ns, 2Es, and 2As and one each of the remaining 5 letters.
The second and third positions can either have two different letters or can both be the same letters.

Case 1: When the two letters are different
We have to choose two different letters from the 8 available different choices.
This can be done in 8 × 7 = 56 ways.

Case 2: When the two letters are same
There are 3 options - the two letters can be Ns or Es or As. Therefore, 3 ways.
Total number of posssibilities = 56 + 3 = 59

Choice A is the correct answer.

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