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It's 8! minus all the unwanted seating arrangements:
So we have to calculate the number of unwanted seating arrangements this way:
Let A = the unwanted case that couple 1 sits together. Let B = the unwanted case that couple 2 sits together. Let C = the unwanted case that couple 3 sits together. Let D = the unwanted case that couple 4 sits together.
So we want N(A or B or C or D)
The "sieve" formula for the answer (sometimes called the "inclusion and exclusion" formula is:
N(A or B or C or D) = N(A) + N(B) + N(C) + N(D) - N(A&B) - N(A&C) - N(A&D) - N(B&C) - N(B&D) - N(C&D) + N(A&B&C) + N(A&B&D) + N(A&C&D) + N(B&C&D) - N(A&B&C&D)
It's easy to see that:
N(A) = N(B) = N(C) = N(D) N(A&B) = N(A&C) = N(A&D) = N(B&C) = N(B&D) = N(C&D) N(A&B&C) = N(A&B&D) = N(A&C&D) = N(B&C&D)
So the "sieve" formula becomes
N(A or B or C or D) = 4N(A) - 6N(A*B) + 4N(A&B&C) - N(A&B&C&D)
We calculate N(A)
We will use (x,y) to mean that a couple sits in seats #x and #y
Choose couple 1's seats 7 ways: (1,2),(2,3),(3,4),(4,5),(5,6),(6,7),(7,8) Choose the way they can sit 2! ways: (husband,wife) or (wife,husband) Seat the other 6 people 6! ways.
That's 7×2!×6! = 10080
N(A&B) = 10080
We calculate N(A&B)
Couples 1 and 2 can use 4 seats in these 15 ways:
{(1,2),(3,4)}, {(1,2),(4,5)}, {(1,2),(5,6)}, {(1,2),(6,7)}, {(1,2),(7,8)}, {(2,3),(4,5)}, {(2,3),(5,6)}, {(2,3),(6,7)}, {(2,3),(7,8)}, {(3,4),(5,6)}, {(3,4),(6,7)}, {(3,4),(7,8)}, {(4,5),(6,7)}, {(4,5),(7,8)}, {(5,6),(7,8)}
Choose whether couple 1 is left of couple 2 or vice-versa in 2! ways: Choose the way couple 1 can sit 2! ways: (husband,wife) or (wife,husband) Choose the way couple 2 can sit 2! ways: (husband,wife) or (wife,husband) Seat the remaining 4 people 4! ways.
That's 15×2!×2!×2!×4! = 2880 ways.
N(A&B) = 2880
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