In how many ways can the letters of the word ‘FAILURE’ be arranged so that the consonants may occupy only odd positions?

Asked by Aaryan | 1 year ago |  59

##### Solution :-

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.

Answered by Sakshi | 1 year ago

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