Question

The number of ways of arranging all the letters of the word "COMBINATIONS" around a circle so that no two vowels come together is 

• A. \( \frac{7!}{(2!)^4} \)
• B. \( \frac{7!}{(2!)^3} \)
• C. \( \frac{^8P_5 \times 6!}{(2!)^3} \)
• D. \( \frac{7! \times ^8P_5}{(2!)^3} \)

Hint:

For circular permutations, always remember that \( n \) objects arranged in a circle have \( (n-1)! \) permutations. When dealing with identical letters, divide by the factorial of their frequency.

Answer 1

The Correct Option is A

Step 1: Identify Consonants and Vowels 
The word "COMBINATIONS" consists of 11 letters: - Vowels: \( O, I, A, I, O \) (5 vowels) - Consonants: \( C, M, B, N, T, N, S \) (7 consonants) To ensure that no two vowels are adjacent, we first arrange the consonants in a circular arrangement. 

Step 2: Arrange Consonants in a Circle 
Since circular permutations of \( n \) distinct objects is given by \( (n-1)! \), the consonants can be arranged in: \[ (7-1)! = 6! \] 

Step 3: Placing Vowels in Gaps 
Once the consonants are placed in a circle, they create 7 gaps. The 5 vowels must be placed in these gaps. The number of ways to choose 5 gaps from 7 is: \[ \text{Ways to place vowels} = ^7P_5 \] Since vowels \( O, I, A, I, O \) include repetitions, we divide by factorials of repeated letters: \[ \frac{7!}{(2!)^4} \] 

Final Answer: \[ \frac{7!}{(2!)^4} \]