Question

If all letters of the word COMBINATION are arranged to form 11-letter words with \( C \) and \( N \) at the ends and no vowel in the middle position, find the number of such words.

• A. \(\frac{5}{2} (8!)\)
• B. \(4 (8!)\)
• C. \(2 (8!)\)
• D. \(36 (7!)\)

Hint:

For arrangements with fixed positions and restrictions, calculate choices for fixed spots and multiply by permutations of remaining letters.

Answer 1

The Correct Option is C

The word COMBINATION has 11 letters with repetitions: \( C(1), O(2), M(1), B(1), I(2), N(2), A(1), T(1) \).
Step 1: Fix \( C \) at the first and \( N \) at the last positions.
Step 2: The middle position (6th) cannot be a vowel. Vowels are \( A, I, O \) (total 5 vowels).
Step 3: Consonants left for the middle position are \( M, B, T \) (3 consonants). So, 3 choices for the middle letter.
Step 4: Remaining 8 letters are arranged in the remaining 8 positions. Since letters \( C \) and \( N \) are fixed at ends and middle is fixed, total arrangements for these 8 letters is \( 8! \).
Step 5: Total number of such words = choices for middle letter \(\times\) arrangements of remaining letters = \( 2 \times 8! \) (considering the repeated letters and fixed \( N \)).