How many words can be formed with the letters of failure when all the vowels should come together ?`?

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

Solution:

The word ‘FAILURE’ has four vowels (E, A, I, U)

The number of consonants is three (F, L, R)

Let’s use the letter C to represent consonants.

1, 3, 5, or 7 are the odd spots.

The consonants can be placed in 4P3 ways in these 4 odd spots.

The remaining three even places (2, 4, 6) will be filled by the four vowels. This can be accomplished in a variety of 4P3 methods. As a result, the total number of words with consonants in odd locations = 4P3 × 4P3.

Using the formula, we can

$ P\text{ }\left( n,\text{ }r \right)\text{ }=\text{ }n!/\left( n-r \right)! $

$ P\text{ }\left( 4,\text{ }3 \right)\text{ }\times \text{ }P\text{ }\left( 4,\text{ }3 \right)\text{ }=\text{ }4!/\left( 4-3 \right)!\text{ }\times \text{ }4!/\left( 4-3 \right)! $

$ =\text{ }4\text{ }\times \text{ }3\text{ }\times \text{ }2\text{ }\times \text{ }1\text{ }\times \text{ }4\text{ }\times \text{ }3\text{ }\times \text{ }2\text{ }\times \text{ }1 $

$ =\text{ }24\text{ }\times \text{ }24 $

$ =\text{ }576 $

As a result, there are 576 different ways to arrange the consonants so that they only appear in odd positions.

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