Why do we subtract [Combinatorics] I asked Here This question and I am still confused. I got that, for at least one group together there are:
$$3 \cdot 9 \cdot \binom{6}{3, 3}$$
But why do we subtract: $3 \cdot 9 \cdot 4$. 
Lets begin with $AAA$ suppose circularly. (We multiply by $3$ in the end for $BBB, CCC$ so not to worry about that). There are $9$ possible places for $AAA$. I saw that: $BCBCBC, CBCBCB$ are two already. Then :$BBCBCC, BCBBCC$, etcc... that is more than $4$.
What am I missing here? 
 A: Consider an arrangement like $AAACBBBCC$, with exactly two blocks of delegates sitting together. The figure
$$3\cdot9\cdot\binom{6}{3,3}\tag{1}$$
counts that arrangement twice, once for the $AAA$ block and once for the $BBB$ block. In fact, every arrangement that has two blocks gets counted twice in $(1)$, so $(1)$ is definitely an overcount. How many of these arrangements are there that we’ve counted twice?
There are $\binom32=3$ ways to choose $2$ countries whose candidates will all sit together. Pick the first of these two countries in alphabetical order: there are $9$ possible positions at the table for their block. That leaves $6$ open seats, and the block for the other chosen country can be placed in any of $4$ positions within that string of $6$ seats. Thus, there are 
$$3\cdot9\cdot4\tag{2}$$ arrangements with two blocks of $3$ delegates sitting together. In $(1)$ we’ve counted each of those arrangements twice, once for each of the two blocks, so we need to subtract $3\cdot9\cdot4$ from $(1)$ to compensate for the original overcounting.
But we’re still not quite done, because of arrangements like $AAABBBCCC$ in which all three countries have their candidates seated together. Each of these arrangements is counted $3$ times in $(1)$, once for each block. Each of these arrangements is also counted $3$ times in $(2)$, once for each pair of blocks: there are $\binom32=3$ pairs of blocks. Thus, in
$$3\cdot9\cdot\binom{6}{3,3}-3\cdot9\cdot4$$
each of these arrangements has been counted $3-3=0$ times. We therefore need to make a final correction: we need to add back in the number of such arrangments. There are $9$ choices of position for country $A$’s block. Once $A$’s block has been seated, there are $2$ possible choices of position for country $B$’s block, and country $C$’s block must then take what’s left. Thus, there are $9\cdot2$ arrangements in which all three sets of delegates are sitting in blocks, and the final correction gives us a total of
$$3\cdot9\cdot\binom{6}{3,3}-3\cdot9\cdot4+9\cdot2$$
arrangements with at least one country’s candidates sitting together.
This is an example of the inclusion-exclusion principle.
A: To directly address the issue you seem to be having: the arrangements with at least two blocks of 3 delegates are the following:


*

*$AAABBBCCC$  

*$AAACBBBCC$  

*$AAACCBBBC$  

*$AAACCCBBB$  


for the specific case of at least the delegates from countries $A$ and $B$ sitting sitting in blocks of 3 (there are also 4 arrangements where the two blocks are the $A$s and $C$s and where they are the $B$s and $C$s).  
From what you have written it seems as if you have forgotten that for the count 3⋅9⋅4 we must have two blocks of 3 delegates. Therefore arrangements like $BCBCBC$ or $CBCBCB$ as you mention in your question do not count as they don't fulfil this requirement, only those 4 listed here do.  
You can also see that 1 and 4 in this list contribute to those cases where at least 3 blocks of 3 delegates sit together. These 2 cases are each counted 3⋅9 times in the figure 3⋅9⋅4 and 3⋅9 times in the figure 3⋅9⋅$\binom{6}{3,3}$ thus cancelling them out. That is why they are added back at the end in the 9⋅2 term.  
A: I believe your specific doubt has been cleared. Here I only want to tell you that when probability has been asked for, whether you take people in a group as distinct and chairs as numbered, or not, it won't affect the answer. You can choose as may be convenient for the problem.
A solution with people taken as distinct, but the chairs as unnumbered, can be seen here
For comparison, the two solutions are summarised below:
[ 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
$\displaystyle 1-\frac{\binom{3}{1}3!6!-\binom{3}{2}4(3!)^{3}+2(3!)^{3}}{8!}=\frac{41}{56}$
