In how many ways can the letters D, I, G, I be arranged so that the two I's are not next to each other?	Select the correct option:	A) 36	B) 48	C) 72	D) 6	E) 128
1 6   2 I   1 G   1 T   Number of ways these letters can be arranged = 4!/2! (2! to account 2 same Is) = 12   Consider 2 Is as 1 entity and thus the number of arrangements for (II)6G = 3! =6   Total allowed cases = 12-6= 
<gadget id="calculator">12-6</gadget>
<output>6</output>
6   6 is the correct answer.
<result>D</result>
In how many ways can the letters D, I, G, I be arranged so that the two I's are not next to each other?