Given: the word is ‘FAILURE.’

To find: number of arrangements so that the consonants occupy only odd positions

Number of vowels in word ‘FAILURE’ = 4(E, A, I, U)

Number of consonants = 3(F, L, R)

Let consonants be denoted by C

Odd positions are 1, 3, 5 or 7

Now, fix the position of consonants like this:

So, arrange these 3 consonants at 4 places

Formula used:

Number of arrangements of n things taken r at a time = P(n, r)

∴ Total number of arrangements of consonants

= the number of arrangements of 4 things taken 3 at a time

= P(4, 3)

= 4!

= 4 × 3 × 2 × 1

= 24

Remaining 3 even places and 1 odd place can be occupied by 4 vowels

So, arrange these vowels at remaining places

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ Total number of arrangements of vowels

= the number of arrangements of 4 things taken all at a time

= P(4, 4)

{∵ 0! = 1}

= 4!

= 4 × 3 × 2 × 1

= 24

Hence, the number of arrangements so that the consonants occupy only odd positions = 24 × 24 = 576