Case 1: The 2nd letter is m and the 3rd letter is e:
The 1st letter may be any of the 4 remaining vowels
Number of possible 3 letter combinations = 4$\begin{bmatrix} a & m & e\\ i & m&e \\ o & m & e\\ u & m & e \end{bmatrix}$
Case 2: The 2nd letter is n and the 3rd letter is e:
The 1st letter may be any of the 5 vowels.
Number of possible 3 letter combinations = 5$\begin{bmatrix} a & n &e \\ e& n & e\\ i& n& e\\ o& n& e\\ u& n& e \end{bmatrix}$
Case 3: The 2nd letter is p and the 3rd letter is e:
The 1st letter will be the same as the 3rd letter.
Number of possible 3 letter combinations = 1 $\begin{bmatrix} epe \end{bmatrix}$
Total number of possible 3 letter combinations
= 4 + 5 + 1 = 10
Hence,Option(C)10 is the correct choice.