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
Is there an error in this question or solution?
Answer : 576
Solution : The given word 'FAILURE' has 3 consonants and 4 vowels, which can be arranged at seven places shown below.
`[""^[1]][""^[2]][""^[3]][""^[4]][""^[5]][""^[6]][""^[7]]`
Now, 3 consonants may be placed at any of the 3 places out of the 4 marked 1, 3, 5, 7.
Number of ways of arranging the consonants `=""^[4]P_[3]=24.`
And, the 4 vowels can be arranged at the remaining 4 places in `
""^[4]P_[4]=24` ways.
Required number of ways `=[24xx24] 576.`
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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Answers
Related The number of arrangements of the letters in the word FAILURE, so that vowels are always coming together isa]576b]575c]570d]none of theseCorrect answer is option 'A'. Can you explain this answer?
There are 4 vovels so we can write statement like this[A,I,U,E]f,L,R now this starement permutation will be 4*3*2*1 but vovels can also rearrange there position so now vovels permutation will be 4*3*2*1 now total arrangements will be 4*3*2*1+4*3*2*1 which is 576
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Answer is 576.
here are seven letters in the word ‘FAILURE’ out of which the consonants are- F, L and R. And the vowels are A,E,I and U. There are total 4 odd positions [1st, 3rd, 5th and 7th] and 3 even positions [2nd, 4th and 6th] to fill.
The constraint is that consonants may occupy only an odd position. So, first we’ll fill the odd positions with the consonants. There are total 4 odd positions and 3 consonants. Since there are only 3 consonants, then at a time, 3 consonants can occupy only 3 odd positions and one will be left out. So, we’ll need to select which three odd positions out of the four available odd positions we will fill first. The number of ways of selecting 3 odd positions out of four are:
This gives us 4 ways to select 3 odd positions out of the available 4 odd positions. Now that we have selected the odd positions to fill, we now need to arrange the consonants in these positions. The number of ways in which we can arrange 3 consonants is equal to 3.Now only the vowels remain to be arranged. There are 4 vowels and 4 vowels can be arranged in 4!.
Now we multiply these results to arrive at the answer: 4∗3!∗4!=576 ways.
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Related The number of arrangements of the letters in the word FAILURE, so that vowels are always coming together isa]576b]575c]570d]none of theseCorrect answer is option 'A'. Can you explain this answer?
The word FAILURE has 7letters 4 wovels and 3 consonants. In total there are 7 letters which are all different.
For all wovels to be together, let's make them a single entity.
So we have 4 entities to be arranged: F, L, R and AIUE
So number of possible arrangements will be 4!
Which is 24.
But the wovels as a group can also be arranged in 4! = 24 ways. [AIUE, AIEU, AUIE, etc.]
So the total number of possible arrangements is 24 * 24 = 576.
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The number of arrangements of the letters in the word FAILURE, so that vowels are always coming together isa]576b]575c]570d]none of theseCorrect answer is option 'A'. Can you explain this answer? for CA Foundation
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There are 4 vovels so we can write statement like this[A,I,U,E]f,L,R now this starement permutation will be 4*3*2*1 but vovels can also rearrange there position so now vovels permutation will be 4*3*2*1 now total arrangements will be 4*3*2*1+4*3*2*1 which is 576