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

Question

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

Ajanta Trivedi · Accepted Answer

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 ${}^{4}{C_{3}}$ 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 = ${}^{4}{C_{3}} \cdot 3! \cdot 4! = 4 * 6 * 24 = 576$ ways

hope this helps you.

cheers!!