How many strings of letters can possibly be formed using the above rules such that the third letter of the string is e?
Explanation:
Case 1: The 2nd letter is m and the 3rd letter is e:
The 1st letter may be any of the 4 remaining vowels (i.e. different from e)
Number of possible 3 letter combinations = 4
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
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 (i.e. ‘epe’)
∴ Total number of possible 3 letter combinations = 4 + 5 + 1 = 10
Hence, option (c).
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