The number of arrangements of the letters of the word "INDEPENDENCE" in which all the vowels always occur together is :
Solutions · 1
Official solution
INDEPENDENCE has letters: I,N,D,E,P,E,N,D,E,N,C,E \to I\times 1, N\times 3, D\times 2, E\times 4, P\times 1, C\times 1.
Vowels: I, E, E, E, E (5 vowels, with E\times 4). Consonants: N, D, P, N, D, N, C (7 consonants: N\times 3, D\times 2, P\times 1, C\times 1).
Treat all vowels as one block, internal arrangement 4!5!=5. Combined with 7 consonants: arrange 8 units with N\times 3, D\times 2: 3!2!8!.
\[\text{Total} = 5\times\frac{8!}{3!2!} = 5\times3360 = 16800\]