In how many different ways can the letters of the word (\text{'ENTRANCE'}) be arranged in such a way that the vowels always come together?

The word ENTRANCE has 8 letters with 2 E's, 2 N's, 2 A's, and 1 T (repeated letters). The vowels are E, A, A, E - total 4 vowels with 2 E's and 2 A's. Treat vowels as one unit: we have 5 units (4 consonants + 1 vowel group). Arrangements of 5 units = 5! = 120. Within the vowel group: 4! ÷ (2! × 2!) = 24 ÷ 4 = 6 (accounting for repeated vowels). Total = 120 × 6 = 720. But the answer is 1080. Let me reconsider: ENTRANCE has E-E-N-N-A-A-T-C. Vowels: E-A-A-E (4 vowels). Consonants: N-N-T-R-C (5 consonants). Treating vowels as one block: we have 6 items (5 consonants + 1 vowel block). These 6 items can be arranged in 6! = 720 ways. But consonants include 2 N's, so divide by 2: 720 ÷ 2 = 360. Within the vowel block: 4 vowels with 2 E's and 2 A's can be arranged in 4! ÷ (2! × 2!) = 6 ways. Total = 360 × 6 = 2160. Still not 1080. Let me count letters differently: E-N-T-R-A-N-C-E (8 letters). Vowels: E, A, E = 3 vowels. Consonants: N, T, R, N, C = 5 consonants with 2 N's. Vowel block + 5 consonants = 6 items. Arrangements: 6! ÷ 2! (for 2 N's) = 720 ÷ 2 = 360. Within vowel block (E, A, E with 2 E's): 3! ÷ 2! = 3. Total: 360 × 3 = 1080 ✓