In how many ways can 6 people, A, B, C, D, E, F be seated in a row such that C and D are not seated next to each other as well as A and B are not seated next to each other?	Answers
A) 384 B) 396 C) 576 D) 624 E) 696
Total # of arrangements of 6 people in a row is 6!;   Now, let's find # of arrangements where C and D as well as A and B ARE seated next to each other. Consider them as a single unit: {AB}, {CD}, E, F. Now, these 4 units can be arranged in 4!, but we should multiply this number by 2*2 as A and B, as well as C and D within their unit can be arranged in two ways: {AB} or {BA}, and {CD} or {DC}. Hence total for this case is 4!*2*2;   So, {# of arrangements}={total}-{restriction}=6!-4!*2*2=624.   Answer: 624.
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In how many ways can 6 people, A, B, C, D, E, F be seated in a row such that C and D are not seated next to each other as well as A and B are not seated next to each other?