Angry couples ....

no. of ways when all three spouses sit together= $\boxed{16}$

no. of ways when two of the three spouses sit together= $\boxed{24}$

no. of ways when only one pair of spouses sit together= $\boxed{48}$

therefore 16+24+48=88

there are total (6-1)! ways... =120

Thus there are 32 ways in which spouses don't sit together...

therefore after simplifying we get 4\15 = 19

Just for convenience we'll assume we are dealing with husband/wife couples. First, put any of the wives in a seat, any seat. Her husband can then be placed either two seats clockwise or counterclockwise, or directly opposite, i.e., $3$ possible seats. Next, consider these $3$ cases one at a time.

1.) two seats clockwise: now place one of the $4$ remaining people in between the already seated couple. The spouse of this individual must then be seated in the chair directly opposite, so as to keep the final couple from being seated together. This final couple can be arranged in two ways, giving us a total of $4*2 = 8$ arrangements in this case.

2.) two seats counterclockwise: same process as in 1.), resulting in $8$ more possible arrangements.

3.) directly opposite: now place one of the remaining $4$ individuals one seat clockwise from the first wife seated. Then one spouse from the remaining couple can be placed one more seat clockwise. The remaining two people, not being spouses, can then be arranged in $2$ ways. Thus we get a total of $4*2*2 = 16$ arrangements in this case.

This gives us a total of $8 + 8 + 16 = 32$ arrangements where no spouses sit next to each other. Since there are $5! = 120$ ways of arranging $6$ people around a circular table without restrictions, (recall rotational symmetry), this results in a desired probability of $\dfrac{32}{120} = \dfrac{4}{15}$ .

This means that $a = 4, b = 15$ and $a + b = \boxed{19}$ .

The probability space is equivalent to arranging 6 points in a circle, and then randomly picking, one by one, a disjoint set of 3 edges joining them (these give the marriages). We are free to choose the order in which we sample edges, and when the first two edges are picked the third is uniquely determined, so we can first pick a segment adjacent to one vertex and then a segment adjacent to one of its neighbors.

Then it's easy to see that there are 3/5 good choices for the first edge. If it's a long diagonal, there are 2/3 good choices for the second edge. If it's a short diagonal, there is only 1/3 good choice for the second edge. This gives $1/5\times 2/3 + 2/5\times 1/3 = 4/15$

This is an old problem, but I thought this was a nice solution.

We will count the number of arrangements in which at least one couple is sitting together.

${3\choose 1}\cdot2^1\cdot4!$ counts each arrangement with one couple together once, with two couples together twice, and with three couples together thrice.

${3\choose 2}\cdot2^2\cdot 3!$ counts each arrangement with two couples together once and with three couples together thrice.

${3\choose 3}\cdot 2^3\cdot 2!$ counts each arrangement with three couples together once.

Therefore, the probability of such an arrangement is $\frac{{3\choose 1}\cdot2^1\cdot4!-{3\choose 2}\cdot2^2\cdot 3!+{3\choose 3}\cdot 2^3\cdot 2!}{5!}=\frac{11}{15}$ . Then the desired probability is $\frac{4}{15}$ , so the answer is $\boxed{19}$ .