Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

In how many different ways can the letters of the word ‘CORPORATION’ be arranged so that the vowels always come together?

seo-qna
Last updated date: 12th Jul 2024
Total views: 346.2k
Views today: 6.45k
Answer
VerifiedVerified
346.2k+ views
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) $ .