In how many ways can the letters of the word ‘FAILURE’ be arranged so that the consonants may occupy only odd positions?
Given: the word is ‘FAILURE.’
To find: number of arrangements so that the consonants occupy only odd positions
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
Now, fix the position of consonants like this:
So, arrange these 3 consonants at 4 places
Formula used:
Number of arrangements of n things taken r at a time = P(n, r)
∴ Total number of arrangements of consonants
= the number of arrangements of 4 things taken 3 at a time
= P(4, 3)
= 4!
= 4 × 3 × 2 × 1
= 24
Remaining 3 even places and 1 odd place can be occupied by 4 vowels
So, arrange these vowels at remaining places
Formula used:
Number of arrangements of n things taken all at a time = P(n, n)
∴ Total number of arrangements of vowels
= the number of arrangements of 4 things taken all at a time
= P(4, 4)
{∵ 0! = 1}
= 4!
= 4 × 3 × 2 × 1
= 24
Hence, the number of arrangements so that the consonants occupy only odd positions = 24 × 24 = 576