The number of arrangements of the letters of the word 'INDEPENDENCE' in which all the vowels always occur together is

The word INDEPENDENCE has 12 letters, of which the vowels are I, E, E, E, E (5 vowels with E repeated 4 times) and the consonants are N, D, P, N, D, N, C (7 consonants with N repeated 3 times and D repeated 2 times).

To count the arrangements in which all the vowels appear together, treat the group of vowels as a single unit. Along with the 7 consonants, this gives 8 units to arrange. Accounting for the repeated letters, the number of ways to arrange these units is $$\frac{8!}{3! \times 2!} = \frac{40320}{6 \times 2} = 3360$$.

Within the vowel group, the 5 letters can be arranged in $$\frac{5!}{4!} = 5$$ ways, since E is repeated 4 times.

Multiplying these yields the total number of arrangements: $$3360 \times 5 = 16800$$.

The correct answer is 16800.