3 x 3 circular arrangement, all distinct In how many ways can we arrange in a circle 3 Angolans, 3 British and 3 Chinese, so that no two people of the same nationality are next to each other? Consider any arrangements that can derive by rotation or flipping of the original arrangement as the same. All people are distinct.
My attempt:
Let’s name the persons A1, A2, A3, B1, B2, B3 and C1, C2, C3.
All possible arrangements are 9!. 
From these we must deduct all arrangements that have at least one pair of adjacent people of the same nationality.
1 pair: All possible ways are 2 x 8!
2 pairs: this can happen either if we have 3 adjacent people of the same nationality or 2 pairs of 2 people. In the 1st case we have 8 different positions for the block of 3 people, so 3x3!x8!
In the 2nd case we have two blocks of 2 people, 2 single people of each nationality and the rest 3 people of the 3rd nationality and 9 positions in total. So 3x (3C2) x 8!
3 pairs: Again 2 cases: either 2 pairs of one nationality and 1 pair of another, or 3 pairs, all of them of different nationality. In the first case we have: 3 x 3! x 2 x 2! x 8!
and in the second case we have 2 x 2 x 2 x 7!
4 pairs: either 2 x 2 x 2 x 3C2  x 6! Or 3 x 6! x 3! x 2! x 2!
5 pairs: 2 pairs of 2 nationalities and 1 pair of a 3rd nationality. So 3C2 x 2! x 2! x 2! x 5!
6 pairs: 2 pairs of each of the 3 nationalities so: either 1 pair + 3 of one nationality + 3 of the other nationality+ 1 single or 3+3+3 so either 2 x 3! x 3! x 4! Or 3! x 3! 
Then by inclusion-exclusion principle.
Anyone can provide a full solution with a result? Also explain the rotation part?
 A: I solved the analogous problem for people arranged in a row here.  However, some of the $37~584$ admissible cases in that problem become inadmissible when we join the ends of the row.  In particular, we have to exclude arrangements in which two people from the same country are at both ends of the row.
For now, let's count arrangements of the letters AAABBBCCC in which both ends of the row are filled by the same letter and no two adjacent letters are identical.  Say the letter C is at both ends of the row.
$$C \square \square \square \square \square \square \square C$$
Then we have to arrange the letters AAABBBC in the middle seven positions so that no two adjacent letters are identical and so that C is not at either end of the row. To do so, we arrange the letters AAABBB so that there is at most one pair of adjacent identical letters.  There are six ways to do this.
$$\color{blue}{ABABAB}, \color{blue}{BABABA}$$
$$\color{red}{ABABBA}, \color{red}{ABBABA}, \color{red}{BAABAB}, \color{red}{BABAAB}$$
In the two alternating arrangements $\color{blue}{ABABAB}$ and $\color{blue}{BABABA}$, the letters A and B are separated, so there are five places in which C can be inserted.
$$\color{blue}{A \square B \square A \square B \square A \square B}$$
In the four arrangements $\color{red}{ABABBA}$, $\color{red}{ABBABA}$, $\color{red}{BAABAB}$, and $\color{red}{BABAAB}$, the C must be placed between the two adjacent identical letters.  Hence, there are 
$$2 \cdot 5 + 4 = 14$$
arrangements of the seven letters AAABBBC in which no two adjacent letters are identical and C does not occupy an end of the row.  Since there are three ways we can choose which letter occupies both ends of the row, there are 
$$3 \cdot 14 = 42$$
arrangements of the letters AAABBBCCC in which no two identical letters are adjacent and the same letter is at both ends of the row.  For each such arrangement, there are $3!$ ways of arranging the representatives of each country in the indicated positions.  Hence, there are 
$$42 \cdot 3!^3 = 9072$$
arrangements of the nine people in which no two representatives of the same country are adjacent and in which representatives from the same country are at both ends of the row.
Hence, if we do not take rotations and reflections into account, the nine people can be arranged in a circle in 
$$37~584 - 9072 = 28~512$$
ways.  Since the same people are next to each other if we reflect ends of the row, there are 
$$\frac{28~512}{2} = 14~256$$
arrangements if reflections are considered equivalent.  Since there are nine rotations that preserve the relative order of the people in each such reflection, there are 
$$\frac{14~256}{9} = 1584$$
admissible arrangements when reflections and rotations are taken into account.  
A: Imagine a circle with the A's placed wide apart, then the B's can be placed in only one way.
$A\quad\quad\quad B\quad\quad\quad A\quad\quad\quad B\quad\quad\quad A\quad\quad\quad B\quad\quad\quad(A)$
There are 6 spots for the C's, $(\bullet)$, as under.
$A\quad\bullet\quad B\quad \bullet\quad A\quad \bullet \quad B \quad \bullet \quad A\quad\bullet\quad B\quad  \bullet\quad (A)$
Thus the C's can be placed in $\binom63 = 20$ ways, as under
$01: ACBCACBAB(A)\quad 02: ACBCABCAB(A)\quad 03: ACBCABACB(A)$
$04: ACBCABABC(A)\quad 05: ACBACBCAB(A)\quad 06: \boxed{ACBACBACB(A)}$
$07: ACBACBABC(A)\quad 08: ACBABCACB(A)\quad 09: ACBABCABC(A)$
$10: ACBABACBC(A)\quad 11: ABCACBCAB(A)\quad 12: ABCACBACB(A)$
$13: ABCACBABC(A)\quad 14: ABCABCACB(A)\quad 15: \boxed{ABCABCABC(A)} $
$16: ABCABACBC(A)\quad 17: ABACBCACB(A)\quad 18: ABACBCABC(A)$
$19: ABACBACBC(A)\quad 20: ABABCACBC(A)$  
Except for $06$ and $15$, which have only one copy under rotation, all the rest have three copies, depending on which $A$ is the starting point, and thus  only generate
$\dfrac{18}3 = 6$ distinct patterns under rotation and reflection.
For example, $01,10,17$ are essentially the same pattern, and $06$ and $15,$ are identical under reflection, thus total number of distinct patterns $= 6 + 1 = 7$
Finally, accounting for distinct persons for each trio of a nation,
number of arrangements $= 7\cdot(3!)^3 = 1512$  
