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

There are 7 letters in the word FAILURE.

We wish to find the total number of arrangements of these 7 letters so that the consonants occupy only odd positions.

There are 3 consonants and 4 odd positions. These 3 consonants can be arranged in the 4 positions in 4! ways.

Now, the remaining 4 vowels can be arranged in the remaining 4 positions in 4! ways.

By fundamental principle of counting:

Total number of arrangements $=4 ! \times 4 !=576$