in how many can the letters of the word'FAILURE' be arranged so that the consonants may occupy only odd positions

the given word is 'FAILURE'

the number of letters = 7

the number of consonants = 3 {F, L, R}

the number of vowels = 4.

the number of odd places = 4 {1st , 3rd , 5th and 7th places}

out of these 4 places consonants can be placed in any of the three places in ways.

and it can be arranged by 3! ways.

4 vowels can be filled in remaining 4 places and can be arranged by 4! ways.

thus the total number of required ways = ways

hope this helps you.

cheers!!

  • 13

Consonants in the word FAILURE: F, I and R.

Vowels in the word FAILURE: A, I, U and E

Given that the consonants in the word FAILURE must occupy odd positions.

There are 4 odd positions. The consonants can be arranged in these 4 positions in 4 x 3! = 24 ways.

The vowels can be arranged in 4! ways.

So total number of ways = 24 x 4! = 576.

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