Combinatorics question about couples sitting around a table 
I have confirmed the solution and posted below.
 A: Divide the 6 seats into 3 groups of 2's, connected to each other, arbitarily. like (12)(34)(56)
The first couple has 3 choices of where they'd like to sit. The 2nd has 2 choices of groups. Once they chose the group of setas, they have two choices (who sits on which side)
so we have $3\times 2 \times 2 \times 2\times 2$. but, we can rotate the table by 60 degrees (i.e. we allocate the table groups differently instead, of seats (12)(34)(56), we allocate it as (23)(45)(61). 
This gives $3\times 2\times 2 \times 2^3/6!= 2/15$
For part (b), first suppose (AA') and (BB') sit together, but not (CC')
A has 6 choices of seats, A' has 2 after A has chosen, B has two choices of seats, as he cannot sit next to A or A', B' has no choices. C has 2 choice of seats.
We coudl have allowed (AA')(CC') and (CC')(BB') to sit together instead and I don't think I double counted
This gives $6\times 2\times 2\times 2\times 3/6!=0.2$
For part (c), again consider two possible configurations (12)(34)(56) and (23)(45)(61)
We have 3 choices of couple we can allow to sit together, with 6 choice of pairs of seats. In each pair, they have 2 choices who sits on which side.
Of the remaining 4 people, 4 choices for the 1st person and no choice for his partner, as she cannot leave 2 adjacent seats. 2 choices for the Husband of the last couple
we have
$6\times 3 \times 2 \times 4\times 2/6!=0.4$
