How many words can be made out of the letters of the word INDEPENDENCE, in which vowels always came together?
A. 16800
B. 16630
C. 1663200
D. None of these

Hint: Here, first see the number of vowels and consonants in the given word then, arrange the vowels and consonants by using permutation formula, then multiply both the arrangements to find the total arrangements of letters with vowels always coming together.

Complete step by step solution:
There are \[12\] letters in the word ‘INDEPENDENCE’ consisting of \[2\] D’s, \[3\] N’s, \[4\] are E’s, and the rest are all distinct.
The word INDEPENDENCE has \[5\] vowels (i.e. I, E, E, E, E) and \[7\] consonants (N, D, P, N, D, N, C)
Arrange\[5\] vowels by using the formula \[\dfrac{{n!}}{{\left( {p! \times q! \times r!} \right)}}\].
[If there are total $n$ letters in which $R$ letters are repeated $p, q$ and $r$ times, number of possible arrangements are given as \[\dfrac{{n!}}{{\left( {p! \times q! \times r!} \right)}}.\]
Therefore, the total arrangement for \[5\] vowels is \[\dfrac{{5!}}{{4!}}\].
Now, arrange the remaining \[8\] letters by using the formula \[\dfrac{{n!}}{{\left( {p! \times q! \times r!} \right)}}\].
\[ = \dfrac{{8!}}{{3!2!}}\]
Hence, total arrangements of letters with vowels always coming together is
$8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1$
$5! = 5 × 4 × 3 × 2 × 1 $
$4! = 4 × 3 × 2 × 1$
$3! = 3 × 2 × 1$
$2! = 2 × 1$
\[ = \dfrac{{8!}}{{3! \times 2!}} \times \dfrac{{5!}}{{4!}} \]
$ = \dfrac{{\left( {8 \times 7 \times 6 \times 5 \times 4 \times 3!} \right) \times \left( {5 \times 4!} \right)}}{{3! \times 4! \times 2}} $
$= 16800 $

$\therefore $ The number of required words = 16800. So, the correct option is (A).

Note:
In these types apply the permutation formula, but keeping in mind the given condition in question, and the repetition of letters. Here in this question, we will multiply both the arrangements of vowels and consonants to find the total arrangements of letters with vowels always coming together.