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 

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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.
Updated On: Mar 15, 2025
  • \( \frac{7!}{(2!)^4} \)
  • \( \frac{7!}{(2!)^3} \)
  • \( \frac{^8P_5 \times 6!}{(2!)^3} \)
  • \( \frac{7! \times ^8P_5}{(2!)^3} \)
     

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The Correct Option is A

Solution and Explanation


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

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