In how many ways can the letters F, E, E, L be arranged so that the two E's are not next to each other? Choose the correct choice:
A) 5
B) 6
C) 72
D) 96
E) 128
1 F   2 E   1 L   Number of ways these letters can be arranged = 4!/2! (2! to account 2 same Es) = 12   Consider 2 Es as 1 entity and thus the number of arrangements for (EE)FL = 3! = 6   Total allowed cases = 12-6= 
<gadget id="calculator">12-6</gadget>
<output>6</output>
6   6 is the correct answer.
<result>B</result>
In how many ways can the letters F, E, E, L be arranged so that the two E's are not next to each other?