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

$\begingroup$

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

$\endgroup$

$\begingroup$

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

1,7531 gold badge13 silver badges15 bronze badges

$\endgroup$

6

$\begingroup$

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

$\endgroup$

2

In how many different ways, can the letters of the words EXTRA be arranged so that the vowels are never together?

1. 168
2. 48
3. 120
4. 72

Answer [Detailed Solution Below]

Option 4 : 72

Calculation:

EXTRA → Total number of words = 5 and total number of vowels = 2

The word EXTRA can be arranged in 5! ways = 120 ways

The word EXTRA can be arranged in such a way that the vowels will be together = 4! × 2!

⇒ [4 × 3 × 2 × 1] × [2 × 1]

⇒ 48 ways

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.

Answer

Verified

Hint: In the given question we are required to find out the number of arrangements of the word ‘CORPORATION’ so that the vowels present in the word always come together. The given question revolves around the concepts of permutations and combinations. We will first stack all the vowels together while arranging the letters of the given word and then arrange the remaining consonants of the word.

Complete step-by-step answer:
So, we are required to find the number of ways in which the letters of the word ‘CORPORATION’ be arranged so that the vowels always come together.
So, we first stack all the vowels present in the word ‘CORPORATION’ separately. So, the vowels present in the word ‘CORPORATION’ are: ‘O’, ‘O’, ‘A’, ‘I’ and ‘O’.
So, we have three O’s , one I and one A.
Now, the remaining consonants in the word ‘CORPORATION’ are: C, R, P, R, T, N.
So, the number of consonants in the word ‘CORPORATION’ is $ 6 $ .
Now, we form a separate bag of the vowels and consider it to be one single entity. Then, we find the arrangements of the letters.
So, the number of entities to be arranged including the bag of vowels is $ 6 + 1 = 7 $ .
Now, the consonant R is repeated twice. So, the number of ways these seven entities can be arranged where one entity is repeated twice are $ \dfrac{{7!}}{{2!}} $ .
Also, there can also be arrangements in the bag of vowels. So, we have five vowels in the bag. But, the vowel O is repeated thrice.
Hence, the number of arrangements in the bag of vowels is $ \dfrac{{5!}}{{3!}} $ .
So, the total number of ways of arranging the letters of the word ‘CORPORATION’ be arranged so that the vowels always come together are $ \dfrac{{7!}}{{2!}} \times \dfrac{{5!}}{{3!}} $ .
Substituting in the values of factorials, we get,
 $ \Rightarrow \dfrac{{5040}}{2} \times \dfrac{{120}}{6} $
Cancelling the common factors in numerator and denominator and simplifying the calculations, we get,
 $ \Rightarrow 50,400 $
So, the correct answer is “ $ \Rightarrow 50,400 $ ”.

Note: One should know about the principle rule of counting or the multiplication rule. Care should be taken while handling the calculations. Calculations should be verified once so as to be sure of the answer. One must know that the number of ways of arranging n things out of which r things are alike is $ \left[ {\dfrac{{n!}}{{r!}}} \right] $ .

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

How many ways can you arrange letters in a word with vowels?

How many ways can the letters be arranged so that all the vowels come together word is impossible?

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