Probabilities in circular arrangements For computing probability for a circular arrangement, it should not matter whether we take people in a group as distinct and chairs as numbered, or not, and we should be able to choose as per our convenience. Such a question is 2011 AIME Problem 12

Nine delegates, three each from three different countries, randomly select chairs at a round table that seats nine people. Let the probability that each delegate sits next to at least one delegate from another country be $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Two solutions (obtained by others, see here and link given therein) were:
[ I ]: Chairs numbered, but only the groups distinct
$\displaystyle 1-\frac{\binom{3}{1}\cdot9{6\choose 3,3}-\binom{3}{2}\cdot9\cdot4 +{3\choose 3}\cdot9\cdot 2}{9\choose 3,3,3}=\frac{41}{56}$
[ II ]: Chairs unnumbered, but people distinct
$\displaystyle 1-\frac{\binom{3}{1}3!6!-\binom{3}{2}4(3!)^{3}+2(3!)^{3}}{8!}=\frac{41}{56}$
I then tried to work out a solution with chairs unnumbered and only groups distinct, i.e. [ I ] modified with chairs unnumbered.
For the numerator in the fractional part, it should suffice to just factor out the 9.
The problem I face is that to get the same answer, the denominator needed is fractional. So where is the error in my working ?
$\displaystyle 1-\frac{\binom{3}{1}{6\choose 3,3}-\binom{3}{2}\cdot4 +{3\choose 3}\cdot 2}{???}=\frac{41}{56}$
 A: To see why this method doesn't work, consider the simpler case where we have 3 people from 2 countries each, and we want to find the probability that each delegate sits next to at least one person from the other country.
$\textbf{1)}$ If we consider the people distinct and the chairs unnumbered, we get
$P(\overline{A}\cap\overline{B})=1-[P(A)+P(B)-P(A\cap B)]=1-\frac{2(3!)^2-(3!)^2}{5!}=1-\frac{36}{120}=\frac{7}{10}$.
$\textbf{2)}$ If we consider the people from each country the same and the chairs numbered, there are 
$\binom{6}{3}=20$ ways to seat the people and 6 ways to seat them with 3 people from the same country together,  
so this gives a probability of $1-\frac{6}{20}=\frac{7}{10}$.

However, if we consider the people from each country the same and the chairs unnumbered, 
there are only 4 possible arrangements of the people, 
and 3 of these arrangements have each delegate next to a delegate  from another country.
The probability is $\textbf{not}$ equal to $\frac{3}{4}$, though, since these possibilities are not equally likely.
[Three of these arrangements each correspond to 6 of the 20 seatings in case $\textbf{2)}$,
while the fourth arrangement only corresponds to 2 of these seatings.]
