There are two German couples, two Japanese couples and one unmarried person 
There are two German couples, two Japanese couples and one unmarried person. If all 9 persons are two be interviewed one by one then the total number of ways of arranging their interviews such that no wife gives an interview before her husband is?

I tried using the string method, but that will be only counting the cases where the wife speaks just after her husband.
There are too many elements to take care of, unlike a similar question which involved only two people, so by symmetry the answer is half of the total number of ways of arranging those $n$ people.
Then how is this one solved
 A: This problem is unusual : you solve a Combinatorics problem by pretending that it is a probability problem.  Normally, the reverse is true, that you solve a probability problem by pretending that it is a Combinatorics problem.
Let $A$ be the event that the German wife precedes the German husband.
Let $B$ be the event that the Japanese wife precedes the Japanese husband.
$A$ and $B$ are independent events, each of which has probability $(1/2).$
Therefore, the answer is $(1/4) \times 9!$.
Edit
Thanks to Aditya Gupta for pointing out my mistake.  For some reason, I thought that there was only 1 Japanese and 1 German couple, rather than 2 of each.
So the correct enumeration is $(1/16) \times 9!$.
A: There are $2$ possible orders of every couple - H W or W H but in this case it is fixed as H W. As they cannot be permuted within, the answer is
$\displaystyle \frac{9!}{ (2!)^4}$ arrangements.

Take an analogy with a string of $9$ characters. Say the string is 1 2 3 4 5 6 7 8 9. There are $9!$ ways to arrange the string.
Now if we have a string 1 2 3 4 5 6 7 8 1, there are $\displaystyle \frac{9!}{2!}$ ways to arrange them as $1$ and $1$ can be arranged in only one way instead of usual $2!$ ways for a pair of distinct characters.
