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

In the word FAILURE, there are $4$ odd position for letters and $3$ even positions.

Now we have $3$ consonants and $4$ places.

No. of ways to arrnage $3$ consonant in $4$ places $={}^4{P}_3=4\times 3\times 2=24$ ways

No. of ways to occupy remaining $4$ vowels $=4!=24$ ways

Total no. ways $=24\times 24=576$ ways