In how many ways can the letters of the word ‘FAILURE’ be arranged so

Question:

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

Solution:

To find: number of words

Condition: consonants occupy odd places

There are total of 7 letters in the word FAILURE.

There are 3 consonants, i.e. $F, L, R$ which are to be arranged in 4 places.

The rest 5 letters can be arranged in 4! Ways.

Formula:

Number of permutations of n distinct objects among $r$ different places, where repetition is not allowed, is

$P(n, r)=n ! /(n-r) !$

Therefore, the total number of words are

$P(4,3) \times 4 !=\frac{4 !}{(4-3) !} \times 4 !=\frac{4 !}{1 !} \times 4 !=\frac{24}{1} \times 24=576$

Hence total number of arrangements is 576.

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