How many arrangements can be formed by the letters of the word vowels if all vowels come together?

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    Permutation is known as the process of organizing the group, body, or numbers in order, selecting the body or numbers from the set, is known as combinations in such a way that the order of the number does not matter.

    In mathematics, permutation is also known as the process of organizing a group in which all the members of a group are arranged into some sequence or order. The process of permuting is known as the repositioning of its components if the group is already arranged. Permutations take place, in almost every area of mathematics. They mostly appear when different commands on certain limited sets are considered.

    Permutation Formula

    In permutation r things are picked from a group of n things without any replacement. In this order of picking matter.

    nPr = (n!)/(n – r)!

    Here,

    n = group size, the total number of things in the group

    r = subset size, the number of things to be selected from the group

    Combination

    A combination is a function of selecting the number from a set, such that (not like permutation) the order of choice doesn’t matter. In smaller cases, it is conceivable to count the number of combinations. The combination is known as the merging of n things taken k at a time without repetition. In combination, the order doesn’t matter you can select the items in any order. To those combinations in which re-occurrence is allowed, the terms k-selection or k-combination with replication are frequently used.

    Combination Formula

    In combination r things are picked from a set of n things and where the order of picking does not matter.

    nCr = n!⁄((n-r)! r!)

    Here,

    n = Number of items in set

    r = Number of things picked from the group

    Solution:

    Vowels are: I,I,O,E

    If all the vowels must come together then treat all the vowels as one super letter, next note the letter ‘S’ repeats so we’d use

    7!/2! = 2520 

    Now count the ways the vowels in the super letter can be arranged, since there are 4 and 1 2-letter(I’i) repeat the super letter of vowels would be arranged in 12 ways i.e., (4!/2!)

    = (7!/2! × 4!/2!) 

    = 2520(12)

    = 30240 ways

    Similar Questions

    Question 1: In how many ways can the letters be arranged so that all the vowels came together word is CORPORATION?

    Solution:

    Vowels are :- O,O,A,I,O

    If all the vowels must come together then treat all the vowels as one super letter, next note the R’r letter repeat so we’d use

    7!/2! = 2520

    Now count the ways the vowels in the super letter can be arranged, since there are 5 and 1 3-letter repeat the super letter of vowels would be arranged in 20 ways i.e., (5!/3!)

    = (7!/2! × 5!/3!)

    = 2520(20)

    = 50400 ways

    Question 2: In how many different ways can the letters of the word ‘MATHEMATICS’ be arranged such that the vowels must always come together?

    Solution:

    Vowels are :- A,A,E,I

    Next, treat the block of vowels like a single letter, let’s just say V for vowel. So then we have MTHMTCSV – 8 letters, but 2 M’s and 2 T’s. So there are

    8!/2!2! = 10,080

    Now count the ways the vowels letter can be arranged, since there are 4 and 1 2-letter repeat the super letter of vowels would be arranged in 12 ways i.e., (4!/2!)

    = (8!/2!2! × 4!/2!)

    = 10,080(12)

    = 120,960 ways

    Question 3: In How many ways the letters of the word RAINBOW be arranged in which vowels are never together?

    Solution:

    Vowels are :- A, I, O  

    Consonants are:- R, N, B, W.

    Arrange all the vowels in between the consonants so that they can not be together. There are 5 total places between the consonants. So, vowels can be organize in 5P3 ways and the four consonants can be organize in 4! ways.

    Therefore, the total arrangements are 5P3 * 4! = 60 * 24 = 1440

    Saurav S.

    asked • 01/05/15

    how many permutations can be formed by the letters of the word 'VOWELS' when

    1>the letters can be placed anywhere

    2>each word begins with E

    3>each word begins with O and ends with I

    4>all vowels come together

    3 Answers By Expert Tutors

    Ralph K. answered • 01/05/15

    Effective Mathematics, Physics and Chemistry Tutor

    1) This is the easiest to do, basic permutations with 6 objects.  The answer is 6! or 6*5*4*3*2*1 = 720

    2) This constrains the problem to a fixed first letter and permutations of 5 objects. The answer is 5! = 5*4*3*2*1 = 120

    3) This constrains the problem to a fixed first and last letter and promotions of 4 objects. The answer is 4! =4*3*2*1 = 24

    4) This is a bit more difficult.  The problem now needs to be viewed as O and E together as a single unit.  Thus we have permutations of 5 objects which as in number 2 above is 120.  This is not the final answer as the vowel grouping has 2 possibilities OE or EO, thus we have 2 times as many permutations which give 2 * 120 = 240.

    1) since you have to pick all placements, you have 6 for the first space and can't reuse it, 5 for the next, then 4,3,2 and 1 which are all being multiplied which is the definition of 6! = 720.

    2) Since I'm specifying the first space must be an E, I have 5 choices for the next, then 4,3,2,1 all being multiplied which is 5! =120.

    3) Since two spaces, the first and last, are specified, you're only picking 4 for the middle 4 spaces, so 4! = 24.

    4) The easiest way of doing this is by picking a specific permutation such as the first 2 are vowels and then the other 4 are consonants. This specific case would be 2*1*4*3*2*1 = 48. Now the vowels could have been in positions 2 and 3....or 3 and 4...or 4 and 5.....or 5 and 6. This makes 5 possibilities for having the vowels together and each specific possibility has 48 ways of occurring, so your total possibilities is 48*5 = 240.

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