Question

The total number of words which can formed using letters of the word ${}^{\prime }FAILUR{E}^{\prime }$ so that consonants always occupy odd places, is-

• A. $144$
• B. $576$
• C. $5040$
• D. None of these

Answer 1

The correct option is B $576$

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 ${}^4{P}_3$ways

$=4\times 3\times 2=24ways$

Remaining must be occupied by vowels in ${}^4{P}_3$

=24

Hence number of ways =$24\times 24=576.$