TS EAMCET · Maths · Permutation Combination
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 \(\frac{7!6!}{(2!)^4}\)
- B \(\frac{7!6!}{(2!)^3}\)
- C \(\frac{{ }^8 \mathbf{P}_5 \times 6!}{(2!)^3}\)
- D \(\frac{7!\times{ }^8 \mathrm{P}_5}{(2!)^3}\)
Answer & Solution
Correct Answer
(A) \(\frac{7!6!}{(2!)^4}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Consonants = C, M, B, N, N, T, S } \\ & \text { Vowels }=0, \mathrm{I}, \mathrm{A}, \mathrm{I}, \mathrm{O}\end{aligned}\) Number of ways consonants can be arranged \(=\frac{(7-1)!}{2!}=\frac{6!}{2!}\) Number of ways vowels can be arranged…
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