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

My way:

Available letters:

M E D I T R A N (8 letters)

E _ _ R

We have 1 combination for E and 1 combination for R, and also we have 8 combinations for the 2nd letter and 7 combinations for the last letter, so:

E 8 7 R = 1 * 8 * 7 * 1 = 56

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

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

anybody plz explain why position 2nd and 3rd not like this : 11*10 = 110 ways !
many 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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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!

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