In how many ways can the letters D, I, S, I, N, O, I, C be arranged so that the three I's are not next to each other? Answers:
A) 36
B) 48
C) 72
D) 96
E) 6000
1 D   3 I   1 S   1 N   1 O   1 C   Number of ways these letters can be arranged = 8!/3! (3! to account 3 same Is) = 6720   Consider 3 Is as 1 entity and thus the number of arrangements for (III)DSNOC = 6! = 720   Total allowed cases = 6720-720= 
<gadget id="calculator">6720-720</gadget>
<output>6_000</output>
6000   6000 is the correct answer.
<result>E</result>
In how many ways can the letters D, I, S, I, N, O, I, C be arranged so that the three I's are not next to each other?