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!!