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Find the different 8 letter arrangements that can be made from the letters of the word DAUGHTER so that all vowels occur together

  • Last Updated : 30 Dec, 2021

In mathematics, permutation relates to the function of ordering all the members of a group into some series or arrangement. In other words, if the group is already directed, then the redirecting of its components is called the process of permuting. Permutations take place, in more or less important ways, in almost every district of mathematics. They frequently appear when different commands on certain limited places are observed.

Permutation

A permutation is known as the process of organizing the group, body, or numbers in order, selecting the or numbers from the set, is known as combinations in such a way that the sequence of the integer does not bother.

Permutation Formula

In permutation, r items are collected from a set of n items without any replacement. In this sequence of collecting matter.

nPr = (n!)/(n – r)!

Here,

n = set dimensions, the total number of object in the set

r = subset dimensions, the number of objects to be choose from the set

Combination

The combination is a way of choosing objects from a group, such that (unlike permutations) the sequence of choosing does not matter. In smaller cases, it is imaginable, to sum up, the number of combinations. Combination refers to the combination of n objects taken k at a time without repetition. To mention combinations in which repetition is allowed, the expressions k-selection or k-combination with repetition are frequently used.

Combination Formula

In combination, r objects are selected from a group of n objects and where the sequence of selecting does not matter.

nCr =n!⁄((n – r)! r!)

Here,

n = Number of objects in group

r = Number of objects selected from the group

Find the different 8 letter arrangements that can be made from the letters of the word DAUGHTER so that all vowels occur together.

Solution:

Total number of letters in DAUGHTER = 8

Vowels in DAUGHTER = A, U, E (vowels are a, e, i, o, u)

Arranging all vowels, since all vowels occur together, they can be AUE, UAE, EAU and so on.

Number of Permutation 3 vowels,

= 3P

= 3!/(3 – 3)!

= 3!/0!

= 3 × 2 × 1 = 6 ways

Arranging 6 letters,

Number we need to arrange = 5 + 1 = 6

Number of permutations of 6 letters,

= 6P6

= 6!/(6 – 6)!

= 6!/0!

= 6 × 5 × 4 × 3 × 2 × 1 = 720

Thus, total number of arrangements = 720 × 6 = 4320

Similar Problems

Question 1: Find the number of different 6 letter arrangements that can be made from the letters of the word FATHER so that all vowels occur together?

Solution:

Total number of letters in FATHER = 6

Vowels in FATHER = A, E (vowels are a, e, i, o, u)

Arranging all vowels, since all vowels occur together, they can be AE, EA and so on.

Number of Permutation 2 vowels,

= 2P2

= 2!/(2 – 2)!

= 2!/0!

= 2 × 1 = 2 ways

Arranging 5 letters,

Number needed to arrange = 4 + 1 = 5

Number of permutations of 5 letters,

= 5P5

= 5!/(5 – 5)!

= 5!/0!

= 5 × 4 × 3 × 2 × 1 = 120

Thus, total number of arrangements = 120 × 2 = 240

Question 2: Find the number of different 8-letter arrangements that can be made from the letters of the word EDUCATION so that all vowels do not occur together?

Solution:

Total number of letters in EDUCATION = 8

Vowels in EDUCATION = E, U, A, I, O (vowels are a, e, i, o, u)

Arranging all vowels

First, calculate when all vowels occur together, they can be EUAIO, UAIOE, AIOUE, IOEUA, OEUAI and so on.

Number of Permutation 5 vowels,

= 5P5

= 5!/(5 – 5)!

= 5!/0!

= 5 × 4 × 3 × 2 × 1 = 120 ways

Arranging 4 letters,

Number we need to arrange = 3 + 1 = 4

Number of permutations of 4 letters,

= 4P4

= 4!/(4 – 4)!

= 4!/0!

= 4 × 3 × 2 × 1 = 24

Thus, total number of arrangements = 120 × 24 = 2,880

So, when all vowels do not occur together, total possible arrangements = 8! – 2880 = 40320 – 2880 = 37440.

Question 3: How many different expressions can be established using the character of the word HARYANA?

Solution:

Total number of characters in HARYANA = 7

The character A repeats 3 times i.e, = 3!

 Number of expression that can be formed

= 7!/3!

= 7 × 6 × 5 × 4 × 3!/3!

​= 840 words

Question 4: What is the number of different expressions beginning and ending with a consonant, which can be completed of the word “EQUATION”? 

Solution:

Total number of characters in EQUATION = 8

8 characters i.e. 3 consonants 5 vowels.

The consonants are to settled 1st and last place and it can be done in 3P2 ways. 

Now 5 vowels and 1 consonant are left i.e. 6 letters which can be organized in 6! ways. Hence the number of expression under given condition is

3P2 × 6! = 6 × 720 = 4320.

Question 5: How many different expressions can be made with the letter of the word ‘ALLAHABAD’?

Solution:

There are total 9 character in the word ‘ALLAHABAD’ in which 4 are ‘A’ s, 2 are ‘L’ and remaining all are definite.

So, the needed number of words

= 9!/4!2! = (9 × 8 × 7 × 6 × 5 × 4!)/4! × 2!

= 7560

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