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