Question

(पाठ-11 दख)
How many arrangements of the letters of the word TAMILNADU can be made if all the vowels
are always together?
(See Lesson-11)
7​

Answer 1

Answer:

Total number of arrangements = 8640

Step-by-step explanation:

Given:

• The word TAMILNADU

To Find:

• Number of arrangements of the letters of the word if the vowels are always together

Solution:

The vowels in the word TAMILNADU are AIAU.

Let us consider the group of these vowels as one letter.

Now number of letters remaining in the word TAMILNADU is,

⇒ 9 - 4 + 1

= 6

Hence number of arrangements of the word = 6!

But the vowels can also be arranged between themselves.

Hence,

The number of ways the vowels can be arranged = 4!/2! ( since letter A repeats two times)

Hence total number of ways of arrangements of the word where all the vowels always occur together = 6! × 4!/2!

⇒ 6 × 5 × 4 × 3 × 2 × 4 × 3

⇒ 8640

Hence there are 8640 different types of arrangements of the word where the vowels are always together.

Answer 2

$\large\underline\blue{\bold{Given \:  Question :-  }}$

• How many arrangements of the letters of the word TAMILNADU can be made if all the vowels are always together?

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$\bf \:\large \red{AηsωeR }$

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$\large\underline\blue{\bold{Formula \: used:- }}$

A permutation is the choice of r things from a set of n things without replacement and where the order matters.

$\boxed{\bf \:^{n}P_r=\dfrac{n!}{(n-r)!}}$

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$\large\underline\purple{\bold{Solution :- }}$

☆In the word TAMILNADU, there are 4 vowels A,A,I,U, in which A repeat twice.

☆So, these vowels can arrange themselves in

$\sf Number\:of\:ways = \: \dfrac{\:^{4} P_4}{\:^{2} P_2} = \dfrac{4!}{2!} = \dfrac{4 \times 3 \times 2!}{2!} = 12$

☆Now, considering these(A,A,I,U) vowels as one set, there are 6 letters ( taking 5 other alphabets) which can be arranged in

$\sf Number\:of\:ways =\:^{6} P_6 = 6! = 720$

☆Hence, by fundamental principle of multiplication, the required number of arrangements are 720 × 12 = 8640.

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