In how many different ways can the letters of the word 'ABUNDANCE' be arranged so that all the vowels always come together?

ABUNDANCE has 8 letters with A repeated twice. Vowels: A, U, A, E (4 letters, with A repeated twice). Treat vowels as one group: we have 5 entities (4 consonants + 1 vowel group). Arrangements = 5! × (4!/2!) = 120 × 12 = 1440. Within vowels: arrange A, U, A, E with A repeated = 4!/2! = 12. Total = 1440. Wait, let me recalculate: consonants (B, N, N, C) - N repeats twice. Total arrangements: (5! × 4!)/(2! × 2!) = (120 × 24)/(2 × 2) = 2880/4 = 720. Hmm, that's not matching either. Let me reconsider: treat AAAE as one group, arrange with B, N, N, C (5 items, N repeats twice) = 5!/2! = 60. Within vowels: A, U, A, E with A repeats = 4!/2! = 12. Total = 60 × 12 = 720. Still not matching options. Let me check: ABUNDANCE letters = A, B, U, N, D, A, N, C, E (9 letters, A twice, N twice). Vowels: A, U, A, E. Consonants: B, N, D, N, C (5 letters, N twice). Total arrangements with vowels together: treat vowels as one block. We have 6 items (5 consonants + 1 vowel block) with N repeating twice among consonants. Arrangement = 6!/2! × 4!/2! = 360 × 12 = 4320. Option B is correct.