How many words can be formed of the letters in the word failure such that the consonants are together and the vowels are also together?

Solution : In the word 'FAILURE' Consonants = F,L,R Vowels = A,I,U,E In the 7 positions No. of even positions = 3 No. of odd positions = 4 No. of ways to fill 3 even positions with 3 consonants `= .^(3)P_(3) = lfloor3 = 6` No. of ways to fill 4 odd positions with 4 vowels `= .^(4)P_(4) = lfloor4 = 24` From multiplication theorem Required permutations `= 6 xx 24 = 144`

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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.
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How many words can be formed of the letters in a word failure such that vowels always come together?
How many words can be formed with the letters of the word failure in which consonants may occupy even places?
How many ways can be letters of the word failure be arranged so that the consonants may occupy only odd place?
How many words can be formed from the letters of the word failure?

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.

Test: Basic Concepts Of Permutations And Combinations- 1

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.

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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How many words can be formed of the letters in a word failure such that vowels always come together?

The number of arrangements of the letters in the word FAILURE, so that vowels are always coming together is. 576.

How many words can be formed with the letters of the word failure in which consonants may occupy even places?

Total no. ways =24×24=576 ways.

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

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

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

Let us cancel the common terms. Therefore, there are 480 words that can be formed using the letters of the word FAILURE so that F is included in each word.