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

To find: number of words

Condition: consonants occupy odd places

There are total of 7 letters in the word FAILURE.

There are 3 consonants, i.e. F, L, R which are to be arranged in 4 places.

The rest 5 letters can be arranged in 4! Ways.

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, the total nuber of words are

P(4,3) ×4! = ×4! = ×4! = ×24 = 576.

Hence total number of arrangements is 576.