If the four letter words (need not be meaningful) are to be formed using the letters from the word "MEDITERRANEAN" such that the first letter is R and the fourth letter is E, then the total number of all such words is:

First, list all the letters in the word “MEDITERRANEAN” together with their individual frequencies:

$$ \begin{aligned} M &: 1,\\ E &: 3,\\ D &: 1,\\ I &: 1,\\ T &: 1,\\ R &: 2,\\ A &: 2,\\ N &: 2. \end{aligned} $$

We must form a four-letter arrangement whose first letter is $$R$$ and whose fourth (last) letter is $$E$$. Because of these fixed positions, we immediately remove one $$R$$ and one $$E$$ from the reservoir of letters. After this removal, the stock that is still available for positions 2 and 3 is

$$ \begin{aligned} M &: 1,\\ E &: 2,\\ D &: 1,\\ I &: 1,\\ T &: 1,\\ R &: 1,\\ A &: 2,\\ N &: 2. \end{aligned} $$

There are now $$11$$ letters in total remaining, spread over $$8$$ distinct kinds. We must choose an ordered pair (second and third positions) from these letters without violating the individual availability of each letter. We treat the two main situations separately.

Case 1: The second and third letters are identical.
For the same letter to fill both positions, at least two copies of that letter must be present in the remaining pool. Looking at the updated counts, only $$E,\,A,$$ and $$N$$ satisfy this requirement (each has 2). Hence the possible identical pairs are

$$EE,\;AA,\;NN.$$ Thus Case 1 contributes $$3$$ arrangements.

Case 2: The second and third letters are different.
We now need an ordered pair of two distinct letters chosen from the 8 kinds available. The usual permutation count for selecting $$r$$ distinct objects in order from $$n$$ distinct objects is

$$ {}^{n}P_r \;=\; \frac{n!}{(n-r)!}. $$

Here, $$n = 8$$ and $$r = 2$$, so

$$ {}^{8}P_2 \;=\; 8 \times 7 \;=\; 56. $$

No further restriction is necessary, because whenever a letter appears only once (such as $$M, D, I, T,$$ or the single remaining $$R$$), we are using it at most once in the ordered pair, which is permissible.

Total count.
Adding the contributions from the two cases, we obtain

$$ \text{Total words} \;=\; 3 \;+\; 56 \;=\; 59. $$

Hence, the correct answer is Option B.