Question:

All the letters of the word are arranged in different possible ways. The number of such arrangements in which no two vowels are adjacent to each other is

Updated On: Jul 6, 2022
  • $360$
  • $144$
  • $72$
  • $54$
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The Correct Option is B

Solution and Explanation

We note that there are $3$ consonants and $3$ vowels $E, A$ and $O$. Since no two vowels have to be together, the possible choice for vowels are the places marked as $ 'X'$ in $XMXCXTX$, these vowels can be arranged in $^{4}P_{3}$ ways, $3$ consonants can be arranged in $ \lfloor 3$ ways. Hence, the required number of ways $ = 3! \times ^{4}P_{3} = 3! \times 4! = 144$.
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Concepts Used:

Permutations and Combinations

Permutation:

Permutation is the method or the act of arranging members of a set into an order or a sequence. 

  • In the process of rearranging the numbers, subsets of sets are created to determine all possible arrangement sequences of a single data point. 
  • A permutation is used in many events of daily life. It is used for a list of data where the data order matters.

Combination:

Combination is the method of forming subsets by selecting data from a larger set in a way that the selection order does not matter.

  • Combination refers to the combination of about n things taken k at a time without any repetition.
  • The combination is used for a group of data where the order of data does not matter.