Question

The number of all possible three letter words that can be formed by choosing three letters from the letters of the word FEBRUARY so that a vowel always occupies the middle place is

• A. 90
• B. 93
• C. 126
• D. 129

Hint:

When dealing with permutations from a set with repeated letters, it's often best to break the problem down into cases: one case where the repeated letters are used, and another where only distinct letters are used.

Answer 1

The Correct Option is B

The letters of the word FEBRUARY are F, E, B, R, U, A, R, Y.
Total 8 letters. Vowels (V) = \{E, U, A\} (3 distinct vowels). Consonants (C) = \{F, B, R, Y, R\} (5 consonants, with R repeated).
The structure of the three-letter word must be $_ V _$.
The middle position must be occupied by a vowel. There are 3 choices for the middle position (E, U, or A).
We can solve this by considering three cases based on the vowel in the middle.
Case 1: The middle letter is E. The remaining letters are \{F, B, R, U, A, R, Y\}. We need to form a 2-letter permutation (for the first and third spots) from this set. The distinct letters available are \{F, B, R, U, A, Y\}. Number of permutations using two distinct letters from these 6 is $^6P_2 = 6 \times 5 = 30$. There is also the permutation using the two R's, which is RR (1 permutation). Total ways for the end positions = $30 + 1 = 31$.
Case 2: The middle letter is U. The remaining letters are \{F, E, B, R, A, R, Y\}. This set is structurally the same as in Case 1 (one pair of R's and 5 other distinct letters). So, the number of ways to fill the end positions is also 31.
Case 3: The middle letter is A. The remaining letters are \{F, E, B, R, U, R, Y\}. Again, this set is structurally the same. So, the number of ways to fill the end positions is also 31.
The total number of possible words is the sum of the ways from all three cases.
Total words = $31 + 31 + 31 = 93$.