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Question
In how many ways can the letters of the word 'FAILURE' be arranged so that the consonants may occupy only odd positions?
Solution
There are 7 letters in the word FAILURE.
We wish to find the total number of arrangements of these 7 letters so that the consonants occupy only odd positions.
There are 3 consonants and 4 odd positions. These 3 consonants can be arranged in the 4 positions in 4! ways.
Now, the remaining 4 vowels can be arranged in the remaining 4 positions in 4! ways.
By fundamental principle of counting:
Total number of arrangements = 4! \[\times\] 4! = 576
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