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

1.	In how many ways can the letters of the word ‘FAILURE’ be arranged so that the consonants may occupy only odd positions?
Answer» 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) x4! = \(\frac{4!}{(4-3)!}\times4!\) = \(\frac{4!}{1!}\times4!\) = \(\frac{24}{1}\times24\) = 576 Hence total number of arrangements is 576.

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

Answer»

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) x4! = \(\frac{4!}{(4-3)!}\times4!\) = \(\frac{4!}{1!}\times4!\) = \(\frac{24}{1}\times24\) = 576

Hence total number of arrangements is 576.