How many ways can the word documentation be arranged so that all the consonants come together?

Method 1: We place the consonants first.

Since ALGEBRA has seven letters, we have seven positions to fill. If we place the B, G, L, and R in that order, we have seven choices where to place the B, six choices where to place the G, five choices where to place the L, and four choices where to place the R. Once the consonants have been placed, there is only one way to arrange the vowels in the remaining positions so that the E appears between the two A's. Hence, there are $$7 \cdot 6 \cdot 5 \cdot 4$$ ways of arranging the letters of the word ALGEBRA so that the relative positions of the vowels are preserved.

Method 2: We place the vowels first.

We choose three of the seven positions for the three vowels. There is only one way to place the vowels in these positions that preserves their relative order. This leaves four positions. We can arrange the four consonants in these positions in $4!$ ways. Hence, the number of ways of arranging the letters of the word ALGEBRA that preserves the relative order of the vowels is $$\binom{7}{3} \cdot 4!$$ Note that this is essentially a rephrasing of @Joffan's solution.

Method 3: We use symmetry.

First, we count the total number of distinguishable arrangements of the letters of the word ALGEBRA. We choose two of the seven positions for the A's. Once the A's have been placed, we can arrange the remaining five distinct letters in the remaining five positions in $5!$ ways. Hence, the number of distinguishable arrangements of the letters of the word ALGEBRA is $$\binom{7}{2} \cdot 5!$$ By symmetry, in one third of these arrangements does the E appear somewhere between the two A's. Hence, the number of arrangements of the word ALGEBRA in which the relative order of the vowels is preserved is $$\frac{1}{3} \binom{7}{2} \cdot 5! = 840$$

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Índice

• In how many ways can the letters of the word IMPOSSIBLE be arranged so that all the vowels come together?
• How many different ways can letter be arranged so that the vowels always come together?
• How many arrangements are there of the word MATHEMATICS if vowels occur together?
• How many ways word arrange can be arranged in which vowels are not together?
• How many ways the word over expand can be arranged so that all vowels come together?

In the below solved problem, every thing is okay, but if we have $4$ consonants then why we are giving $5!$? and is this a combination problem? how to distinguish?

Question: In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?

Answer: The word 'OPTICAL' contains $7$ different letters. When the vowels OIA are always together, they can be supposed to form one letter. Then, we have to arrange the letters PTCL [OIA]. Now, $5$ letters can be arranged in $5! = 120$ ways. The vowels [OIA] can be arranged among themselves in $3! = 6$ ways. Required number of ways $= [120*6] = 720$.

asked Jun 26, 2014 at 16:40

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There are $4$ consonants and $1$ group of vowels, so there are $5$ elements to permute. Yes, this is a combinatorial problem because we are counting the number of possibilities that satisfy certain conditions.

answered Jun 26, 2014 at 16:45

fahrbachfahrbach

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For this problem I use the multinomial coefficient. We have 7 letters in total 3 of which are vowels and 4 are consonants

The number of vowels combinations are 3! and the number of consonants combinations are 4! [I've used the multinomial coefficient] So by the end we've 3!*4! ways

answered Dec 14, 2020 at 22:37

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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

In how many ways can the letters of the word IMPOSSIBLE be arranged so that all the vowels come together?

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

How many different ways can letter be arranged so that the vowels always come together?

The number of ways the word TRAINER can be arranged so that the vowels always come together are 360.

How many arrangements are there of the word MATHEMATICS if vowels occur together?

In the word 'MATHEMATICS', we treat the vowels AEAI as one letter. Thus, we have MTHMTCS [AEAI]. Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different. ∴ Number of ways of arranging these letters = 8!

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

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

How many ways the word over expand can be arranged so that all vowels come together?

The word EXTRA can be arranged in such a way that the vowels will be together = 4! × 2! The letters of the words EXTRA be arranged so that the vowels are never together = [120 - 48] = 72 ways. ∴ The letters of the words EXTRA be arranged so that the vowels are never together in 72 ways.

∴ The number of ways of arranging these letter is 64800.

Since, the vowels have to be together, we can say that we have to arrange the groups [C], [R], [P], [R], [T], [N] and [OOAIO] among themselves. Considering the objects of the same type, this can be done in. 2 ! = 2520 ways.

[b] 120960 There are 11 letters in the word “MATHEMATICS” out of which 4 are vowels and the rest 7 are consonants. Corresponding to each of these permutations, the 4 vowels can be arranged among themselves in 4! 2!

Total numbers of letters, we have to arrange = 5+1=6. Total number of arrangements when all vowels occur together=720×6=4320.