How many different ways can the letters of the word combine be arranged so that vowels always come together?

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

Consider a seven-letter word formed by mixing up the letters in the word COMBINE. How many ways can you do this if no vowel is isolated between two consonants?
(eg. EBMCION and MOIENCB are acceptable, but BEMCNIO is not)

The final answer is given to be $1872$.

We are being asked to find the numeric value of (Total number of seven-letter words) - (Number of words such that a vowel is isolated between two consonants). The word "COMBINE" has $4$ consonants and $3$ vowels. To compute the number of constructed words with an isolated vowel between two consonants, I tried the following:

1st attempt:

We may generalize the case, representing consonants as $1$s and vowels as $0$s. Now, we are finding the total number of seven-digit binary strings with four $1$s and three $0$s, containing the substring $101$. There are $\frac{4!}{2!2!} = 6$ strings that can be formed from the remaining two $0$s and $1$s. We may insert the substring $101$ into any position in these 6 strings. We have five available positions, because there are four digits. Therefore, there are $6 \times 5=30$ seven-digit binary strings with four $1$s and three $0$s, containing the substring $101$.

Since each vowel and consonant is unique in the word COMBINE, there are $30 \times 4!\times3!=4320$ seven-letter words formed from COMBINE that contain an isolated vowel between two consonants, if we account for permutations of the vowels and consonants. However, $7!-4320=720 \neq 1872$, meaning my answer is incorrect.

2nd attempt:

Since there are three vowels, for there to be an isolated vowel between two consonants, either all vowels are separated, or two vowels may be grouped up together while the remaining vowel is isolated (there is no other case).

Case when all vowels are isolated:
First, we arrange the consonants, Then, we choose three out of five possible places to place the vowels (.C.C.C.C., where C denotes a consonant and the periods represent possible places to put a vowel). We finally account for the arrangement of the vowels. $$ P(4,4) \times C(5,3) \times P(3,3) = 1440.$$

Case when one vowel is isolated, while the remaining two are a pair:
First, we arrange the consonants. We then choose a vowel. Then, we choose a position out of three possible places to put it (C.C.C.C). Then, we choose a position out of four possible places to place the pair of the remaining vowels (.CVC.C.C. just as a possible example). Finally, we account for the arrangement of the vowels in the pair. $$ P(4,4) \times C(3,1) \times C(3,1) \times C(4,1) \times P(2,2) = 1728.$$ $7!-(1440+1728) = 1872$, which confirms that this is the correct answer.

What is wrong with my first attempt? Where did I make my error?

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