Q:

2011 AIME Problem 12, probability round table

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 $m+n$.

I have been searching around math stackexchange and I saw this difficult question, so I decided to try to work it out myself.
Question: I am numbering the chairs from (top to last) as $1, 2, 3, 4, 5, 6, 7, 8, 9$. 
Now I am stuck, how many ways are there TOTAL to arrange the people? 
We want $P = 1 - P(\overbrace{\text{at least one block is the same or all three.}}^{x}) = 1 - P(x).$
The denominator is: $\binom{9}{3, 3, 3}$.This is the number of ways to make three groups of three objects in each. 
Next: How many ways are there to make one block identical eg. $AAA$?
But for each $AAA$ there are three cases. Fix $a$ on chair 1 then there are possibilities: $AAa, aAA, AaA$. Thus for the $A$ there are: $3\binom{6}{3, 3}$ ways to arrange at least one block. Same for $B, C$ hence, $9\binom{6}{3, 3}$ total. 
But in this we counted the $AAABBBCCC$ arrangements several times. For each $AAA$ there are two possible $AAABBBCCC$ sequences. So subtract $2(9) = 18$. 
Hence:
$P = 1 - \frac{9\binom{6}{3, 3} - 18}{\binom{9}{3, 3, 3}} = \frac{253}{280}$ 
WRONG answer. I am confused, what I did wrong?

A:

In your setup we have $\binom9{3,3,3}=1680$ possibilities in total. 
Let $n(A)$ denote the number of possibilities such that you have a block $AAA$. Likewise we use notations $n(B)$ and $n(C)$. Then $n(A)=n(B)=n(C)=9\binom6{3,3}=180$. 
Let $n(A\cap B)$ denote the number of possibilities s.t. there is a block $AAA$ and a block $BBB$. Then $n(A\cap B)=n(A\cap C)=n(B\cap C)=9\times4=36$
Let $n(A\cap B\cap C)$ denote the number of possibilities s.t. there is a block $AAA$, a block $BBB$ and a block $CCC$. Then $n(A\cap B\cap C)=9\times2=18$
Finally let $n(A\cup B\cup C)$ denote the number of possibilitis such that there is block $AAA$ or a block $BBB$ or a block $CCC$. 
With inclusion/exclusion we find:
$$n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C)=3\times180-3\times36+18=450$$
So there is probability of $\frac{450}{1680}=\frac{15}{56}$ that there is at least one block and a probability of $\frac{41}{56}$ that there is no block.
The numbers $41$ and $56$ are coprime, so the final answer is $41+56=97$.
Hints for another setup can be found here.

