<s>[INST] How many ways are there to distribute 12 distinguishable balls into 6 distinguishable bins? How many ways are there to distribute 12 distinguishable balls into 6 distinguishable bins so that 2 balls are placed in each bin ? 

I solved it in the following way =>
$C(12,2) * C(10,2) * C(8,2) * C(6,2) * C(4,2) * C(2,2)$ and got the correct answer.

Can I apply Sterling formula of 2nd kind here ?
I want to know how Sterling formula of 2nd kind is applied here ?
 [/INST] The number I think of immediately is the multinomial coefficient
$$
\begin{align}
\binom{12}{2,2,2,2,2,2}
&=\frac{12!}{2!\,2!\,2!\,2!\,2!\,2!}\\[6pt]
&=7484400
\end{align}
$$
which can be computed from binomial coefficients as you have done.

However, if you want to use Stirling Numbers of the Second Kind and avoid multinomial coefficients, we can compute the number of partitions of $12$ items with no singleton using Inclusion-Exclusion: The number of ways to partition a set of $12$ items into $6$ subsets with $k$ specific singletons is
$$\newcommand{\stirtwo}[2]{\left\{#1\atop#2\right\}}
\overbrace{\ \ \ \ \binom{12}{k}\ \ \ \ }^{\substack{\text{number of ways to}\\\text{choose the singletons}}}\overbrace{\ \ \stirtwo{12-k}{6-k}\ \ }^{\substack{\text{number of ways to}\\\text{partition the rest}}}
$$
By Inclusion-Exclusion, there are then
$$
\sum_{k=1}^6(-1)^{k-1}\binom{12}{k}\stirtwo{12-k}{6-k}
$$
ways to partition $12$ items into 6 subsets with at least one singleton. Since there are $\stirtwo{12}{6}$ ways to partition $12$ items into 6 subsets, we have
$$
\begin{align}
&\stirtwo{12}{6}-\sum_{k=1}^6(-1)^{k-1}\binom{12}{k}\stirtwo{12-k}{6-k}\\
&=\sum_{k=0}^6(-1)^k\binom{12}{k}\stirtwo{12-k}{6-k}
\end{align}
$$
ways to partition $12$ items into 6 subsets with no singletons; that is, into $6$ pairs. However, this does not count the orderings of these $6$ pairs, so we need to multiply this number by $6!$. That is,
$$
\begin{align}
&6!\sum_{k=0}^6(-1)^k\binom{12}{k}\stirtwo{12-k}{6-k}\\[3pt]
&=6!\left[{\textstyle\binom{12}{0}\stirtwo{12}{6}-\binom{12}{1}\stirtwo{11}{5}+\binom{12}{2}\stirtwo{10}{4}-\binom{12}{3}\stirtwo{9}{3}+\binom{12}{4}\stirtwo{8}{2}-\binom{12}{5}\stirtwo{7}{1}+\binom{12}{6}\stirtwo{6}{0}}\right]\\[9pt]
&=720\,[{\small1\cdot1323652-12\cdot246730+66\cdot34105-220\cdot3025+495\cdot127-792\cdot1+924\cdot0}]\\[9pt]
&=720\cdot10395\\[10pt]
&=7484400
\end{align}
$$
</s>