How many ways can the letters of the word failure be arranged so that the consonants may occupy only?

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

There are 7 letters in the word FAILURE.
We wish to find the total number of arrangements of these 7 letters so that the consonants occupy only odd positions.
There are 3 consonants and 4 odd positions. These 3 consonants can be arranged in the 4 positions in 4! ways.
Now, the remaining 4 vowels can be arranged in the remaining 4 positions in 4! ways.

By fundamental principle of counting:
Total number of arrangements = 4! \[\times\] 4! = 576

Concept: Factorial N (N!) Permutations and Combinations

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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.

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

Solution

There are 4 vowels and 3 consonants in the word 'FAILURE' We have to arrange 7 letters in a row such that consonants occupy odd places. There are 4 odd places (1, 3, 5, 7). There 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. Hence, the total number of words in which consonants occupy odd places = 4P3×4P3 = 4!(4−3)!×4!(4−3)! = 4×3×2×1×4×3×2×1 = 24×24 = 576.

How many ways can the letters of the word failure be arranged so that consonants occupy odd position?

How many different ways can the letters of the word failure be arranged so that the vowels occupy only the odd positions?

How many words can be formed with the letters of the word failure?

How many ways the letters of the word failure can be arranged with the condition that the four vowels are always together?