1. Count repeated letters in INDEPENDENCE

   The word INDEPENDENCE has 12 letters:

   $I, N, D, E, P, E, N, D, E, N, C, E$

   Their frequencies are:

   • $E = 4$
   • $N = 3$
   • $D = 2$
   • $I = 1$
   • $P = 1$
   • $C = 1$

2. Identify vowels and consonants

   Vowels are: $I, E, E, E, E$ So total vowels $= 5$.

   Consonants are: $N, N, N, D, D, P, C$ So total consonants $= 7$.

3. Treat all vowels as one block

   Since all vowels must always occur together, treat the 5 vowels as a single unit.

   Then we arrange the following 8 units:

   • 1 vowel block
   • $N, N, N$
   • $D, D$
   • $P$
   • $C$

   So the number of arrangements of these 8 units is: $\frac{8!}{3!\,2!}$

4. Arrange the vowels within the block

   Inside the vowel block, the letters are: $I, E, E, E, E$

   Number of distinct arrangements: $\frac{5!}{4!} = 5$

5. Total number of required arrangements

   Therefore, $\text{Total} = \frac{8!}{3!\,2!} \times 5$

   Compute: $\frac{8!}{3!\,2!} = \frac{40320}{6\cdot 2} = \frac{40320}{12} = 3360$

   Hence, $3360 \times 5 = 16800$

6. Check options

   • A: $16800$ ✅
   • B: $14800$ ❌
   • C: $18000$ ❌
   • D: $33600$ ❌

Therefore, the correct answer is: $\boxed{16800}$