Question

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

Answer 1

In the word “FAILURE’ there are 4 odd position for letters and three even positions.

Now we have 3 consonants and 4 places. Hence they can be arranged in 4P3=4∗3∗2=244P3=4∗3∗2=24 ways

Remaining must be occupied by vowels in 4!ways=244!ways=24

Hence number of ways =24∗24=576