If you want to find the number of ways to arrange men and women so that we have no men or women sitting consecutively we get
$$2\cdot5!\cdot5!$$
Since There $5!$ ways to arrange the males skipping a seat in between and $5!$ ways to arrange the females in the remaining $5$ spots. Then there are $2$ ways to start the row, either male or female.
Also if you're trying to find the number of ways to only group men together and women together (MMMMMWWWWW and WWWWWMMMMM) the number is the same, $2\cdot5!\cdot5!$
Label the couples A, B, C so that the six people are A$_{1}$, A$_{2}$, B$_{1}$, B$_{2}$, C$_{1}$ and $C_{2}$.
Now, consider your 6 seats: _ _ _ _ _ _
There are 6 choices for who can sit in the first seat; without loss of generality, say it's A$_{1}$.
There are 4 choices for who can sit in the second seat (anyone else but A$_{2}$); without loss of generality, let's say this is B$_{1}$.
Our table now looks like this: A$_{1}$ B$_{1}$ _ _ _ _
For the 3rd seat, we just can't have B$_{2}$, so we can either have A$_{2}$ or one of the C's; I'll split this into 2 cases.
Case 1: A$_{2}$ sits in the third seat.
A$_{1}$ B$_{1}$ A$_{2}$ _ _ _
Now the fourth seat can't be B$_{2}$, otherwise the C's will sit next to each other in the last two seats. So the fourth seat must be one of the C's; there are 2 choices for this. Then the fifth seat must be B$_{2}$ and the sixth must be the other C.
Thus there are $6*4*1*2*1*1 = 48$ ways for Case 1 to happen.
Case 2: One of the C's sits in the third seat; there are 2 possibilities for this. Let's say it's C$_{1}$
A$_{1}$ B$_{1}$ C$_{1}$ _ _ _
Then there are 2 choices for the fourth seat: A$_{2}$ or B$_{2}$. I'll treat these as subcases.
Case 2.1: A$_{2}$ sits fourth.
A$_{1}$ B$_{1}$ C$_{1}$ A$_{2}$ _ _
For the fifth seat, we have 2 choices, and then 1 for the sixth.
Thus $6*4*2*1*2*1 = 96$ possibilities for this case.
Case 2.2: B$_{2}$ sits fourth.
A$_{1}$ B$_{1}$ C$_{1}$ B$_{2}$ _ _
For the fifth seat, there is only one possibility: A$_{2}$, since otherwise A$_{2}$ would be in the sixth seat, which is adjacent to the first (the table is circular). Of course there is only one possibility for the sixth.
Thus there are $6*4*2*1*1*1 = 48$ ways for Case 2.2 to happen.
There are then $48+96+48 = 192$ possible seating arrangements. The probability of this happening is $\frac{192}{6!} = \frac{192}{720} = \frac{4}{15} \approx 0.267$
Best Answer
There are two cases, (1) where we disregard everything about a person except their nationality, and (2) where we don't. Start with (1).
Label the nationalities A,B,C. Suppose the first two seats are AB. Then the third seat can be A or C. Suppose it is A. It is not hard to see that there are just 9 possibilities:
ABAB followed by CACBC or CBCAC; ABACA followed by BCBC or CBCB; ABACBACBC; or ABACBC followed by ABC, ACB, BAC or BCA.
If the third seat is C, then we have ABCAB, ABCAC, ABCBA or ABCBC. Each of these will give the same number of possibilities because each has used up two instances each of two letters and one of the third.
ABCABA must be followed by CBC. ABCABC must be followed by one of ABC, ACB, BAC, BCA. That is 5 ways. So 4 x 5 = 20 in all for ABC...
So we have a total of 29 ways for seating beginning AB... . There are 6 ways it could begin (AB, BA, AC, CA, BC, CB). So that gives a grand total of 174 for case (1). For case (2) we must multiply by $6^3$ (there are 6 ways of sitting three nationals in 3 given seats), so we get 37584.