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: You got the total probability right.
Now regarding the numerator. (I would like to note that in your solution, you regarded the delegates from each country to be indistinguishable). We need to find the number of ways that each delegate sits next to at least one delegate from another country
The number of ways to make one block identical is actually not what you have. Suppose you have AAA as one block. Then the other six delegates can be arranged in $\binom{6}{3, 3}$ ways. However, you do not arrange the 3 delegates inside the AAA block. Instead, you can put the AAA block in any of the 9 chairs. So we have $9\times \binom{6}{3, 3}$ ways to have an AAA block. So the number of ways to have an identical block is $3 \times 9\times \binom{6}{3, 3}=27\times 20.$
We now need the number of ways for candidates from two countries to each sit together. We can do this in $3\times9\times4=27\times4$ ways. (I believe this is the part you didn't calculate)
Finally we need the number of ways that the candidates from all countries sit together. This can be done in $9\times2=18$ ways, one clockwise and the other anticlockwise.
So the total number of unwanted arrangements is $27\times20-27\times4+18.$
This should lead you to the right answer (I don't know the answer because the AOPS site seems to be down, but I got $m+n$ to be $97.$)
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.
