Question -
Answer -
Given:
The word ‘FAILURE’
Number of vowels inword ‘FAILURE’ = 4(E, A, I, U)
Number of consonants =3(F, L, R)
Let consonants bedenoted by C
Odd positions are 1,3, 5 or 7
The consonants can bearranged in these 4 odd places in 4P3 ways.
Remaining 3 evenplaces (2, 4, 6) are to be occupied by the 4 vowels. This can be done in 4P3 ways.
So, the total numberof words in which consonants occupy odd places = 4P3 × 4P3
By using the formula,
P (n, r) = n!/(n – r)!
P (4, 3) × P (4, 3) =4!/(4 – 3)! × 4!/(4 – 3)!
= 4 × 3 × 2 × 1 × 4 ×3 × 2 × 1
= 24 × 24
= 576
Hence, the number ofarrangements so that the consonants occupy only odd positions is 576.