Given:
The word ‘FAILURE’
Number of vowels in word ‘FAILURE’ = 4(E, A, I, U)
Number of consonants = 3(F, L, R)
Let consonants be denoted by C
Odd positions are 1, 3, 5 or 7
The consonants can be arranged in these 4 odd places in 4P3 ways.
Remaining 3 even places (2, 4, 6) are to be occupied by the 4 vowels. This can be done in 4P3 ways.
So, the total number of words in which consonants occupy odd places = 4P3 × 4P3
= 4 × 3 × 2 × 1 × 4 × 3 × 2 × 1
= 24 × 24
= 576
Hence, the number of arrangements so that the consonants occupy only odd positions is 576.
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