How many different arrangements of A, B, C, D, E AND F are possible where A is not adjacent to B and D is not adjacent to E?
Answers:  A) 96  B) 48  C) 32  D) 144  E) 1
So you have A,B,C,144,E,F   five possible ways to arrange it = 6! = 720   AB cannot be adjacents neither 144E:   (AB),C,144,E,F = 5! =120 x 2(because it can be (AB) or (BA) ) = 240   A,B,C,(144E) = 5! = 120 x 2 = 240   To get rid of the double counting:   (AB),C,(144E),F = 4! x 4 = 96   (Four possible ways to arrange it)   720 -240-240+96 = 144 possible ways   144
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How many different arrangements of A, B, C, D, E AND F are possible where A is not adjacent to B and D is not adjacent to E?