Question

The total number of words which can be formed using the letters of the word ‘FAILURE’ so that consonants always occupy odd places, is-
$A.{\text{ 144}} \\\ {\text{B}}{\text{. 576}} \\\ C.{\text{ 5040}} \\\ {\text{D}}{\text{. None of these}} \\\$

Answer 1

Hint- In this question we have used the concept of permutation. At first we write the no. of vowels and then consonants after that we have to write the no. of positions for odd and even. After that, using conditions we will find no. of ways to arrange letters.
Complete step-by-step answer:
There are 4 vowels (a, e, i, u,) and 3 consonants (f, l, r) in the word FAILURE.
There are 4 odd positions (1, 3, 5, 7) and 3 even positions (2, 4, 6)
The given condition is that the consonants can occupy only odd positions, i.e., the three consonants are to occupy three positions out of the four odd positions available.
For the first consonant, there are four choices, the second has three choices and the last has two choices. Now the 4 vowels are to be arranged in 4 places.
The no. of ways 4 letters can be arranged in 4 places $= 4!$
Therefore, the total no. of ways = $4 \times 3 \times 2\left( {4!} \right)$
$= 24 \times \left( {24} \right) = 576$
Hence, (B) is the correct option.
Note- In this particular question one may wonder why 4! is being multiplied and not added to $4 \times 3 \times 2$, it is because the consonants are arranged first ‘and’ then vowels are arranged. A simple way to remember this as follows when ‘and’ is used we should multiply and whenever ‘or’ is used, we should add. This is where we can make mistakes.