Placing different colors of indistinguishable balls around a circle 
$3n$ indistinguishable balls are coloured with $n$ colours so that each colour is to be used exactly three times. In how many ways these coloured balls can be placed around a circle so that $3$ balls with same colour never appears side by side?

Using inclusion-exclusion principle , i found that $$\sum_{i=0}^{n}\binom{n}{i}(-1)^i\frac{[3n-(2i+1)]!}{3!\times(n-i)}$$
However , i am not sure about my answer.. I suspect that i am doing overcounting and Polya must have been used. What do you think ?
 A: Your work is correct except for two mistakes. The final answer is
$$
\left[\sum_{i=0}^n (-1)^i\binom{n}{i}\frac{(3n-2i-1)!}{\color{red}{(3!)^{n-i}}}\right]+\color{red}{\frac{2\cdot n!}{3n}}\tag{$*$}
$$
You had $(3!)\times (n-i)$ instead of $(3!)^{n-i}$, so it appears that you just confused multiplication with exponentiation. The summand of $\color{red}{{2\cdot n!}/{(3n)}}$ comes from the Polya counting method, see equation $(3)$ in the full explanation below.

For each $i\in \{0,1,\dots,n\}$, let
$$
L_i=\text{# linear arrangements of $3n$ balls, where colors $1,\dots,i$ are all together}\\
C_i=\text{# circular arrangements of $3n$ balls, where colors $1,\dots,i$ are all together}
$$
The numbers $C_i$ are what we are after, since the inclusion exclusion gives
$$
\text{# circular arrangements where no color is altogether}=\sum_{i=0}^n (-1)^i\binom niC_i\tag1
$$
However, $L_i$ is easier to count, so let us start with that. We have to imagine that for each color between $1$ and $i$, the three balls in that color are grouped together. The total number of objects to arranged is then $3(n-i)+i=3n-2i$, and there are $(n-i)$ groups of three identical objects in that total, so
$$
L_i=\frac{(3n-2i)!}{(3!)^{n-i}},\qquad 0\le i\le n\tag2
$$
Now, how to we account for circular arrangements? Let us start by computing $C_0$. For this, we use the Polya counting method.  The symmetry group is $\mathbb Z/3n\mathbb Z$, and we need to take the average over that group of the number of fixed points of each rotation. For the trivial identity permutation, the number of fixed points is the number of all linear arrangements, $L_0$. It turns out the only nontrivial symmetries with fixed points correspond to rotation by $120^\circ$ and $240^\circ$, that is rotation by $n$ spots or by $2n$ spots. Here, fixed points correspond to linear arrangements of $n$ balls in different colors repeated three times in a row; there are $n!$ ways to do this, for each of the two symmetries. This proves that
$$
C_0=\frac{1}{3n}\left(L_0+\color{red}{2\times n!}\right)\tag3
$$
This explains the other part of the final answer you were missing.
Finally, we just need the value of $C_i$ for $1\le i\le n$. Here, the symmetry group is $\Bbb Z/(3n-2i)\Bbb Z$, since there are only $3n-2i$ objects. Applying Polya's method is even easier; since we have at least one special group of three balls joined together, there are no fixed points of any nontrivial symmetry. This means that
$$
C_i=\frac1{3n-2i}\cdot L_i=\frac1{(3n-2i)}\frac{(3n-2i)!}{(3!)^{n-i}}=\frac{(3n-2i-1)!}{(3!)^{n-i}}\qquad 1\le i\le n\tag4
$$
Combining $(1),(2),(3)$ and $(4)$ yields $(*)$.

As a sanity check, when $n=2$, my formula yields
$$
\left[\frac{5!}{3!^2}-2\cdot \frac{3!}{3!}+\frac{1!}{1} \right]+\frac{2\cdot 2}{3\cdot 2}=\boxed{3}
$$
Sure enough, there are three arrangements in this case, illustrated below:
  ⬤ ◯            ◯ ⬤            ⬤ ◯      
◯     ⬤        ◯     ⬤        ⬤     ◯   
  ⬤ ◯            ⬤ ◯            ◯ ⬤   

A: Here is my contribution. For PIE the underlying poset consists of nodes
$Q \subseteq [n]$ ordered by set inclusion which represent configurations where the colors in  $Q$
are adjacent, plus possibly more adjacent ones. The weight on the
configurations represented at each  node of the poset is $(-1)^{|Q|}$.
Now for a configuration that has  exactly the set $P$ of adjacent colors
(monochrome three-sequence) with $P$ not empty summing over all nodes the
total contribution is
$$\sum_{Q\subseteq P} (-1)^{|Q|} =
\sum_{q=0}^{|P|} {|P|\choose q} (-1)^q
= (1-1)^{|P|} = 0,$$
so we get not contribution from these nodes, as required. On the other
hand for configurations with no adjacent three-tuples these are only
included in the node $Q=\emptyset$ for a total weight of
$(-1)^{|\emptyset|} = 1,$ and they get counted exactly once, also as
required.
Now we need to count the number of configurations represented by a node
$Q$. This is done with the Polya Enumeration Theorem. The cycle index
for an $m$-necklace is
$$Z(C_m) = \frac{1}{m} \sum_{d|m} \varphi(d) a_d^{m/d}.$$
Let $|Q|=q\ge 1.$ Apply PET to get
$$[R_1 R_2 \cdots R_{q} S_1^3 S_2^3 \cdots S_{n-q}^3]
Z(C_{3n-2q};
R_1 + R_2 + \cdots + R_q + S_1 + S_2 + \cdots + S_{n-q}).$$
Here we see by inspection that only $d=1$ can possibly contribute and
we get
$$\frac{1}{3n-2q}
[R_1 R_2 \cdots R_{q} S_1^3 S_2^3 \cdots S_{n-q}^3]
(R_1 + R_2 + \cdots + R_q + S_1 + S_2 + \cdots + S_{n-q})^{3n-2q}
\\ = \frac{1}{3n-2q} \frac{(3n-2q)!}{3!^{n-q}}.$$
Now when $q=0$ we obtain
$$[S_1^3 S_2^3 \cdots S_{n}^3]
Z(C_{3n};
S_1 + S_2 + \cdots + S_{n}).$$
Here we get contributions from $d=1$ and $d=3$. The first is
$$\frac{1}{3n} \frac{(3n)!}{3!^n}$$
and the second is
$$\frac{2}{3n} [S_1^3 S_2^3 \cdots S_{n}^3]
(S_1^3 + S_2^3 + \cdots + S_{n}^3)^n
\\ = \frac{2}{3n} [S_1 S_2 \cdots S_{n}]
(S_1 + S_2 + \cdots + S_{n})^n
= \frac{2\times n!}{3n}.$$
Substitute the first case into PIE and add the second to obtain
$$\bbox[5px,border:2px solid #00A000]{
\frac{2\times n!}{3n}+
\sum_{q=0}^n {n\choose q} (-1)^q \frac{(3n-2q-1)!}{3!^{n-q}}}$$
and we confirm the accepted answer.
