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All the Letters of the Word 'Eamcot' Are Arranged in Different Possible Ways. Find the Number of Arrangements in Which No Two Vowels Are Adjacent to Each Other.

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Question

All the letters of the word 'EAMCOT' are arranged in different possible ways. Find the number of arrangements in which no two vowels are adjacent to each other.

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Solution

We note that, there are 3 consonants M, C, T and 3 vowels E, A, O.

Since, no two vowels have to be together, the possible choice for volwels are the blank spaces

\[_M_C_T_\]

These vowels can be arranged in 4P3 ways.
3 consonants can be arranged in 3! ways.

Hence, the required numbers of ways = 3! × 4P3 = 144 ways.

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Factorial N (N!) Permutations and Combinations
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Chapter 16: Permutations - Exercise 16.3 [Page 29]

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R.D. Sharma Mathematics [English] Class 11
Chapter 16 Permutations
Exercise 16.3 | Q 32 | Page 29

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