Simple probaility

Thenumber of ways of taking seats without restriction
= (5 - 1)!
= 24

To fulfill the requirement :
A and B first take their seats ((2 - 1)!).
Then, C and D fill in the two spaces between A and B (P(2,2)).
Lastly, E fill in 1 space of the 4 spaces (P(4,1)).

The number of ways to fulfill the requirement
= (2 - 1)! × P(2,2) × P(4,1)
= 1 × 2! × (4!/3!)
= 8

The required probability
= 8/24
= 1/3

2014-08-28 03:02:36 補充：
Alternative method :

Consider that case when 1 or 2 couples sit next to each other :
Each couple form a group. The two groups take their seat ((2 - 1)!).
The internal permutation of the two groups is P(2,2) × P(2,2)
Then E choose a seat from the 4 spaces between 2 person (P(4,1))

2014-08-28 03:02:56 補充：
No. of ways in this case
= (2 - 1)! × P(2,2) × P(2,2) × P(4,1)
= 1 × 2! × 2! × 4!/3
= 16

No. of ways that fulfills the requirement
= 24 - 16
= 8

The required probability
= 8/24
= 1/3

Let the seats numbers be 1,2,3,4,5. Consider E to take the seat first, no. 1, say.
So the couple A,B can only take (2,4) or (3,5) as C,D should take the other 2.
So there only 2*2!*2! ways, that is, 8 ways only.