Q56 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics
The number of ways of arranging all the letters of the word "COMBINATIONS" around a circle so that no two vowels come together is
- Correct answerA$\frac{7!6!}{(2!)^4}$
- B$\frac{7!6!}{(2!)^3}$
- C$\frac{{ }^8 \mathbf{P}_5 \times 6!}{(2!)^3}$
- D$\frac{7!\times{ }^8 \mathrm{P}_5}{(2!)^3}$
Solution
$\begin{aligned} & \text { Consonants = C, M, B, N, N, T, S } \\ & \text { Vowels }=0, \mathrm{I}, \mathrm{A}, \mathrm{I}, \mathrm{O}\end{aligned}$
Number of ways consonants can be arranged $=\frac{(7-1)!}{2!}=\frac{6!}{2!}$ Number of ways vowels can be arranged $=\frac{5!}{2!2!}$ Number of ways so that no vowel come together ${ }^7 \mathrm{C}_5$ Total number of ways $={ }^7 \mathrm{C}_5 \times \frac{6!}{2!} \times \frac{5!}{2!2!}=\frac{7!6!}{(2!)^4}$
Number of ways consonants can be arranged $=\frac{(7-1)!}{2!}=\frac{6!}{2!}$ Number of ways vowels can be arranged $=\frac{5!}{2!2!}$ Number of ways so that no vowel come together ${ }^7 \mathrm{C}_5$ Total number of ways $={ }^7 \mathrm{C}_5 \times \frac{6!}{2!} \times \frac{5!}{2!2!}=\frac{7!6!}{(2!)^4}$
Hence, the correct answer is option: 'A'.