In how many ways can the letters of the word FAILURE be arranged so that the consonants may occupy only odd positions?

Answer

576

Explanation

The given word 'FAILURE' has 3 consonants and 4 vowels, which can be arranged at seven places shown below.
`(""^(1))(""^(2))(""^(3))(""^(4))(""^(5))(""^(6))(""^(7))`
Now, 3 consonants may be placed at any of the 3 places out of the 4 marked 1, 3, 5, 7.
Number of ways of arranging the consonants `=""^(4)P_(3)=24.`
And, the 4 vowels can be arranged at the remaining 4 places in ` ""^(4)P_(4)=24` ways.
Required number of ways `=(24xx24) 576.`