Official solution
Here total no of different letters present are,
(1) One M
(2) Three E (E E E)
(3) One D
(4) One I
(5) One T
(6) Two R (R R)
(7) Two A (A A)
(8) Two N (N N)
In the four letter word first letter is R and last letter is E.
\[ \therefore \] Word is = R \_ \_ E
Now remaining letters are,
M, EE, D, I, T, R, AA, NN
Those 2 empty places can be filled with identical letters [EE, AA, NN] in 3 ways.
Or two empty places can be filled with distinct letters [M, E, D, I, T, R, A, N] in \[{}^8{C_2} \times 2!\] ways.
\[ \therefore \] Total no of words = 3 + \[{}^8{C_2} \times 2!\] = 59