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 n objects arranged in a circle have (n1)! (n-1)! permutations. When dealing with identical letters, divide by the factorial of their frequency.
Updated On: Mar 15, 2025
  • 7!(2!)4 \frac{7!}{(2!)^4}
  • 7!(2!)3 \frac{7!}{(2!)^3}
  • 8P5×6!(2!)3 \frac{^8P_5 \times 6!}{(2!)^3}
  • 7!×8P5(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 O, I, A, I, O (5 vowels) - Consonants: C,M,B,N,T,N,S 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 n distinct objects is given by (n1)! (n-1)! , the consonants can be arranged in: (71)!=6! (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: Ways to place vowels=7P5 \text{Ways to place vowels} = ^7P_5 Since vowels O,I,A,I,O O, I, A, I, O include repetitions, we divide by factorials of repeated letters: 7!(2!)4 \frac{7!}{(2!)^4}  

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

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