The number of arrangements of the letters of the word "INDEPENDENCE" in which all the vowels…

There are $5$ vowels in the given word which are $4E\text{'}s \&\hspace{0.17em}1I$.

Since they have to always occur together we take them as a single object $E E E E I$ for the time being.

This single object together with $7$ remaining object will account for $8$ objects.

There $8$ objects in which there are $3N\text{'}s \&\hspace{0.17em}2D\text{'}s$ can be arrangement in $\frac{8!}{3!2!}$ ways.

Corresponding to each of their arrangements the $5$ vowels $E,E,E,E \& I$ which can be arranged in $\frac{5!}{4!}$

Hence, required number of arrangements.

$=\frac{8!}{3!2!}\times \frac{5!}{4!}=16800$

Hence this is the correct option.