<s>[INST] Suppose that you have a set of 12 colored balls, two each of six different colors $C_{1}$ through $C_{6}$... Suppose that you have a set of 12 colored balls, two each of six different colors $C_{1}$ through $C_{6}$. Find the number of six-ball combinations if balls of the same color are considered identical. 
Using the Inclusion-Exclusion Principle and $\binom{n}{m}_{R}=\binom{n+m-1}{m}$ where m is the number of balls and n is the number of boxes (or in this case the length of the combination), I should he able to solve this problem.
However, I have not idea how to solving this problems. Can someone please help me? Thank you
 [/INST] Generating Function Approach
$$
\begin{align}
\left[x^6\right]\left(1+x+x^2\right)^6
&=\left[x^6\right]\left(\frac{1-x^3}{1-x}\right)^6\\
&=\left[x^6\right]\sum_{k=0}^6(-1)^k\binom{6}{k}x^{3k}\sum_{j=0}^\infty(-1)^j\binom{-6}{j}x^j\\
&=\sum_{k=0}^2(-1)^k\binom{6}{k}(-1)^{6-3k}\binom{-6}{6-3k}\\
&=\sum_{k=0}^2(-1)^k\binom{6}{k}\binom{11-3k}{5}\\
&=\binom{6}{0}\binom{11}{5}-\binom{6}{1}\binom{8}{5}+\binom{6}{2}\binom{5}{5}\\[9pt]
&=462-336+15\\[15pt]
&=141
\end{align}
$$

Casing on the Number of Pairs
Choose $k$ colors to have two balls of each of those colors and one of the rest
$$
\begin{align}
\sum_{k=0}^3\overbrace{\ \ \ \binom{6}{k}\ \ \ }^{\substack{\text{number of}\\\text{ways to choose}\\\text{paired colors}}}\overbrace{\binom{6-k}{6-2k}}^{\substack{\text{number of}\\\text{ways to choose}\\\text{singleton colors}}}
&=\binom{6}{0}\binom{6}{6}+\binom{6}{1}\binom{5}{4}+\binom{6}{2}\binom{4}{2}+\binom{6}{3}\binom{3}{0}\\
&=1+30+90+20\\[9pt]
&=141
\end{align}
$$
</s>