SOLUTION: four married couple went to attend a concert in how many ways can they be arranged:-(a) if they sit together (b)if there are no restrictions (c)if each wife does not sit next to he

We can choose the couple to sit on the far left in 4 ways.
We can choose the way they sit 2 ways. (husband left of wife or wife left of husband).

We can choose the couple to sit next to the far left in 3 ways.
We can choose the way they sit 2 ways. (husband left of wife or wife left of husband).

We can choose the couple to sit next to the far right in 2 ways.
We can choose the way they sit 2 ways. (husband left of wife or wife left of husband)

We can choose the couple to sit on the far right in just 1 way.
We can choose the way they sit 2 ways. (husband left of wife or wife left of husband).

That's  4×2×3×2×2×2×1×2 = 4!×24 = 384 ways

Choose the person to sit in the 1st seat 8 ways.
Choose the person to sit in the 2nd seat 7 ways.
Choose the person to sit in the 3rd seat 6 ways.
Choose the person to sit in the 4th seat 5 ways.
Choose the person to sit in the 5th seat 4 ways.
Choose the person to sit in the 6th seat 3 ways.
Choose the person to sit in the 7th seat 2 ways.
Choose the person to sit in the 8th seat 1 ways.

That's 8×7×6×5×4×3×2×1 = 8! = 40320 ways.

---------------------------------------------------

Wow!  This is the only hard one of the 3.

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

---------------------------

We calculate N(A&B&C) 

Couples 1, 2 and 3 can use 6 seats in these 10 ways:
 
{(1,2),(3,4),(5,6)}, {(1,2),(3,4),(6,7)}, {(1,2),(3,4),(7,8)},
{(1,2),(4,5),(6,7)}, {(1,2),(4,5),(7,8)}, {(1,2),(5,6),(7,8)}, 
{(2,3),(4,5),(6,7)}, {(2,3),(4,5),(7,8)}, {(2,3),(5,6),(7,8)},
{(3,4),(5,6),(7,8)}

We can choose the order of the 3 couples is 3! or 6 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)
Choose the way couple 3 can sit 2! ways: (husband,wife) or (wife,husband)
Seat the remaining 2 people 2! ways.

That's 10×3!×2!×2!×2!×2! = 960 ways.

N(A&B&C) = 960

--------------------------------------

We calculate N(A&B&C&D)
(They use all 8 seats)
Choose the order of the 4 couples 4! = 24 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)
Choose the way couple 3 can sit 2! ways: (husband,wife) or (wife,husband)
Choose the way couple 4 can sit 2! ways: (husband,wife) or (wife,husband)

That's 4!×2!×2!×2!×2! = 384

----------------------------

Substitute in "sieve" formula to get the number of unwanted seating
arrangements:

N(A or B or C or D) = 4N(A) - 6N(A&B) + 4N(A&B&C) - N(A&B&C&D)

N(A or B or C or D) = 4(10080) - 6(2880) + 4(960) - 384 = 26496

---------------------------

Subtract from 8!:

8! - 26496 = 40320 - 26496 = 13824.

Answer: 13824

Edwin