Questions

How many ways can the letters of the word rainbow be arranged so that only two vowels always remain together?

How many ways can the letters of the word rainbow be arranged so that only two vowels always remain together?

Hence total arrangements = 4×72=288. =1440 ways.

How many words each of 3 vowels and 2 consonants can be formed from the letters of the word daughter?

Therefore, 30 words can be formed from the letters of the word DAUGHTER each containing 2 vowels and 3 consonants.

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How many words of 3 consonants and 2 vowels can be formed?

Aptitude :: Permutation and Combination – Discussion Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed? = 210. Number of groups, each having 3 consonants and 2 vowels = 210. Each group contains 5 letters.

How many words can be formed with the letters of the word rainbow in which A is always before I and I is always before O?

Our answer = 6!

How many ways can the letters in the word rainbow be arranged?

These arrangements are when the two vowels are always together. If you subtract from total arrangements 7! you get the same ans… 5,040 ways.

How many arrangements each consisting of 2 vowels & 2 consonants can be made out of the letters of the word devastation?

Hence, the number of words each consisting of two vowels and two consonants which can be made out of the letters of the word ‘DEVASTATION’ is 1638.

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How many ways can the letters of the word rainbow be arranged?

This result 4320 is overcounting since it is only considering the fact that all vowels come together only and not the case where only two vowels can also come together also, as the question says vowels are never together, and not that all the vowels are not together only. This is why the correct answer is 1440.

How many words can be formed from the letters of the word rainbow?

93 words can be made from the letters in the word rainbow.

How many ways can the letters of the word rainbow be arranged in which vowels are never together?

How many ways can the word indicator be arranged if the vowels and consonants are arranged in alternate positions?

So, with vowels at even places, total number of arrangements will be 6 x 6=36. Similarly, with vowels at odd places, the number of arrangements will also be 36. Hence, total number of arrangements with vowels and consonants at alternative places will be 36+36=72.