Question

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

Answer 1

There are 4 vowels and 3 consonants in the word 'FAILURE'
We have to arrange 7 letters in a row such that consonants occupy odd places. There are 4 odd places (1, 3, 5, 7). There consonants can be arranged in these 4 odd places in $4_{P_3}$ ways.
Remaining 3 even places (2, 4, 6) are to be occupied by the 4 vowels. This can be done in $4_{P_3}$ ways.
Hence, the total number of words in which consonants occupy odd places = $4_{P_3}\times 4_{P_3}$
= $\frac{4!}{(4-3)!}\times \frac{4!}{(4-3)!}$
= $4\times 3\times 2\times 1\times 4\times 3\times 2\times 1$
= $24\times 24$
= 576.