How many arrangements of the word Bengali can be made if the vowels are never together?

PermutationCombinationPermutation(Arrangement)1. Letters1.The number of new words that can be formed by rearranging the letters of the word 'ALIVE' is -.A. 24B. 23`C.119D. 120

2.Find the number of ways of arranging the letters of the word "MATERIAL" such that the vowels in the word are to come together?

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3.In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?

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4.How many arrangements of the letters of the word ‘BENGALI’ can be made, if the vowels are never together?

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5.The number of arrangements that can be made with the letters of the word “MEADOWS” so that vowels occupy the even placesA. 720B.144C. 120D. 36

6.In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy odd positions?

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7.In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?

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8.How many arrangements can be made of the letters of the word “ASSASSINATION”? In how many of them are the vowels always together?2!!

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2. Linear objects9.In how many ways can 10 books be arranged on a shelf such that a particular pair of books should always be together?A. 9! × 2!B. 9!C. 10! × 2!D. 10!

10.In how many ways can 10 books be arranged on a shelf such that a particular pair of books will never be together?

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11.In how many ways can 11 persons be arranged in a row such that 3 particular persons should always be together?

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12.In how many ways can a team of 5 persons be formed out of a total of 10 persons such that two particular persons should be included?

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13.In how many ways can 10 software engineers and 10 civil engineers be seated in a row so that they are positioned alternatively?A. 2×(10!)2B. 2×10! × 11!C. 10! × 11!D.(10!)2

14.Kiran has 8 black balls and 8 white balls. In how many ways can he arrange these balls in a row so that balls of different colors alternate?2

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15.A company has 11 software and 7 civil engineers. In how many ways can they be seated in a row so that no two civil engineers will sit together?

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Answer

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  • How many arrangements of the letters of the word KEYBOARD can be made if the vowels are to occupy odd places in the word arrangement?
  • How many ways can the letters of the word KEYBOARD can arranged such way that vowels always come together?
  • How many arrangements are there if all the vowels are together?
  • How many arrangements of the letters of the word Bengali can be made if the vowels are to occupy only odd places?

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Hint: In this particular question use the concept that the number of ways to arrange r different objects from n different objects is equal to ${}^n{P_r}$, and the number of ways to arrange n different objects is equal to n!, and use the concept that if we want some letters not together so first find out all the words with these letters together then subtract these words from the total possible words, so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given word:
‘BENGALI’
As we see that all the letters in the given word are different and the number of letters are 7.
As we all know there are 5 vowels present in the English alphabets which are given as A, E, I, O and U.
Out of these alphabets the number of alphabets present in the given word BENGALI are E, A and I.
So in the given word there are 3 vowels and 4 consonants.
Now the total number of words which are possible form the letters of given word BENGALI = 7!
 $\left( i \right)$ If all vowels are never together.
Consider that all the vowels are together so consider three vowels present in the given word as one letter.
So the arrangements of vowels internally = 3!
So there are 5 letters in the word so the number of arrangements = 5!
So the total number of words possible when all the vowels are together = \[\left( {5! \times 3!} \right)\]
So the total number of words such that no vowel is together is the difference of total number of words from the given word and the total number of words when all the vowels are together.
Therefore, the total number of words such that no vowel is together = \[7! - \left( {5! \times 3!} \right)\]
Now simplify we have,
$ \Rightarrow \left( {7.6.5!} \right) - \left( {6.5!} \right) = 6.5!\left( {7 - 1} \right) = 36\left( {5!} \right) = 36\left( {120} \right) = 4320$ Words.
$\left( {ii} \right)$ If the vowels are to occupy only odd places.
As we see that there are 7 letters in the given word. So there are 4 odd places which is ${1^{st}},{3^{rd}},{5^{th}}{\text{ and }}{{\text{7}}^{th}}$ places.
As there are three vowels so the number of ways to arrange them in 4 odd places = ${}^4{P_3}$.
So three vowels are filled in the odd places, so there are 4 letters remaining and there are 4 places remaining so the number of ways to fill these 4 letters in 4 places = 4!
So the total number of words such that vowels comes in odd places = ${}^4{P_3} \times 4!$
Now as we know that ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ so use this property in the above equation we have,
So the total number of ways such that vowels comes in odd places = $\dfrac{{4!}}{{\left( {4 - 3} \right)!}} \times 4!$
$ \Rightarrow \dfrac{{4!}}{{\left( {4 - 3} \right)!}} \times 4! = 4! \times 4! = 24\left( {24} \right) = 576$ Words.
So the total number of words such that vowels come in odd places are 576.
Hence option (b) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the formula of permutation which is stated above and always recall that in a English alphabets there are 5 vowels which are given as, A, E, I, O, and U, so according to this find out the number of vowels present in the given word.

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There are five stations on a railway line. What is the number of different journey tickets that are required for railway authorities?

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6 1616

Q:

If each of the vowels in the word 'MEAT' is kept unchanged and each of the consonants is replaced by the previous letter in the English alphabet, how many four-lettered meaningful words can be formed with the new letters, using each letter only once in each word?

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Q:

In a bag containing red, green and pink tokens, the ratio of red to green tokens was 5 : 12 while the ratio of pink to red tokens was 7 : 15. What was the ratio of green to pink tokens?

A) 25 : 28 B) 36 : 7
C) 8 :25 D) 12 : 7

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Q:

The ratio of the members of blue to red balls in abag is constant. When there were 44 red balls, the number of blue balls was 36. If the number of blue balls is 54, how many red balls will be in the bag?

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Q:

If each vowel of the word NICELY is changed to the next letter in the English alphabetical series and each consonant is changed to the previous letter in the English alphabetical series and then the alphabets are arranged in alphabetical order, which of the following will be fourth from the left?

Answer & Explanation Answer:

Explanation:

Thus after arranging the letters as per English alphabetical series; we get; Thus 4th letter from the left end will be K.

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Q:

How many such pair of letters are there in the word ‘TROUBLED’ which have as many letters between them in the word as they have between them in the English alphabet?

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Q:

If it is possible to make a meaningful word with the first, the seventh, the ninth and the tenth letters of the word RECREATIONAL, using each letter only once, which of the following will be the third letter of the word? If more than one such word can be formed, give ‘X’ as the answer. If no such word can be formed, give ‘Z’ as the answer.

Answer & Explanation Answer: D) R

Explanation:

The first, the seventh, the ninth and the tenth letters of the word RECREATIONAL are R, T, O and N respectively. Meaningful word from these letters is only TORN. The third letter of the word is ‘R’.

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6 3114

Q:

16 persons are participated in a party. In how many differentways can they host the seat in a circular table, if the 2particular persons are to be seated on either side of the host?

A) 16! × 2 B) 14! × 2
C) 18! × 2 D) 14!

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6 4099

How many arrangements of the letters of the word KEYBOARD can be made if the vowels are to occupy odd places in the word arrangement?

Total number of such possible words is 2880 Total ways of allocating vowels = 4 * 3!

How many ways can the letters of the word KEYBOARD can arranged such way that vowels always come together?

This is Expert Verified Answer 9.

How many arrangements are there if all the vowels are together?

Mathematic can be arranged in 453,600 different ways if it is ten letters and only use each letter once. Assuming all vowels will be together 15,120 arrangements.

How many arrangements of the letters of the word Bengali can be made if the vowels are to occupy only odd places?

=9P4×4! ×3! =24×24=576.

How many arrangements of the letters of the word Bengali If vowels are never together?

=9P4×4! ×3! =24×24=576. Was this answer helpful?

How many arrangements of the letters of the word Bengali can be made if the vowels are to occupy only odd places?

There are 7 letters in the word Bengali of these 3 are vowels and 4 consonants. There are 4 odd places and 3 even places. 3 vowels can occupy 4 odd places in 4P3 ways and 4 constants can be arranged in 4P4 ways. There are five stations on a railway line. ... .

How many words are there in which vowels are never together?

number of arrangements in which the vowels do not come together =5040−1440=3600 ways.

How many ways word arrange can be arranged in which vowels are not together?

Hence, the answer is 36.