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Above is the original question. The correct answer is in green that is 59.

I have chosen option 3 that is $\frac{11!}{(2!)^3}$ because I thought that there are 13 alphabets in the word "mediterranean" and now we are fixing two alphabets so we are left with 11 alphabets and out of this "n", "a" and "e" occur two times so the total number of permutations will be $\frac{11!}{(2!)^3}$.

It is clear that my answer is wrong but I can't understand what is wrong and how to solve to get the correct answer.

I can only think that the alphabets which we are fixing occur more than one time so this may be the cause of my mistake but how?

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  • $\begingroup$ The problem to your suggestion is that you calculate all the permutations, as if we had to fill in $11$ gaps, not just $2$ (as we want here). $\endgroup$ – thanasissdr Apr 22 '16 at 3:34
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It looks like they are looking for words that have exactly 4 letters. Since there are 8 different letters that can be chosen (m, e, d, i, t, r, n, and a), there are $8 \cdot 7 = 56$ possible combinations where the second and third letter are different. There are also 3 possibilities where the 2nd and 3rd letter are the same (ee, rr, or nn).

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    $\begingroup$ "..... the same (ee, aa, or nn)." $\endgroup$ – true blue anil Apr 22 '16 at 4:08

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