Answer

Verified

373.5k+ 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) $ .

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Which are the Top 10 Largest Countries of the World?

Give 10 examples for herbs , shrubs , climbers , creepers

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

How do you graph the function fx 4x class 9 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Change the following sentences into negative and interrogative class 10 english CBSE