If each digit in the set of A = (6,7,8,9) is exactly used once, in how many ways can the digits be arranged such that the 6 and 7 are not adjacent?
Pick.
A) 12
B) 15
C) 18
D) 20
E) 22
The answer should be = total arrangements - arrangements when 2 and 5 are together.   = 4! - 3! * 2! = 24 -12= 
<gadget id="calculator">24 -12</gadget>
<output>12</output>
12   take (6,7),8,9 here there are 3 entities whose number of arrangements are 3! and 2! is the arrangement between 3 and 4   12nswer is 12
<result>A</result>
If each digit in the set of A = (6,7,8,9) is exactly used once, in how many ways can the digits be arranged such that the 6 and 7 are not adjacent?