Question:
In how many ways can the letters of the word ‘FAILURE’ be arranged so that the consonants may occupy only odd positions?
Solution:
To find: number of words
Condition: consonants occupy odd places
There are total of 7 letters in the word FAILURE.
There are 3 consonants, i.e. $F, L, R$ which are to be arranged in 4 places.
The rest 5 letters can be arranged in 4! Ways.
Formula:
Number of permutations of n distinct objects among $r$ different places, where repetition is not allowed, is
$P(n, r)=n ! /(n-r) !$
Therefore, the total number of words are
$P(4,3) \times 4 !=\frac{4 !}{(4-3) !} \times 4 !=\frac{4 !}{1 !} \times 4 !=\frac{24}{1} \times 24=576$
Hence total number of arrangements is 576.