<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] Applying $S_2$ here will be quite convoluted.
As explained in another question of yours,
The number of ways  of getting exactly k different values when you throw n fair r-sided dice is 
$\binom{r}{k}\cdot S_2(n,k)\cdot k! $
so to get exactly $6$ faces in $12$ throws of a normal die, $\binom66\cdot S_2(12,6)\cdot6!$ ways,
but this would include $11$ patterns, viz.
$\boxed 7\boxed 1\boxed 1\boxed 1\boxed 1\boxed 1\;\;\boxed 6\boxed 2\boxed 1\boxed 1\boxed 1\boxed 1\;\;\boxed 5\boxed 3\boxed 1\boxed 1\boxed 1\boxed 1\;\; \boxed 5\boxed 2\boxed 2\boxed 1\boxed 1\boxed 1\;\;\boxed 4\boxed 4\boxed 1\boxed 1\boxed 1\boxed 1$
$\boxed 4\boxed 3\boxed 2\boxed 1\boxed 1\boxed 1\;\;\boxed 4\boxed 2\boxed 2\boxed 2\boxed 1\boxed 1\;\;\boxed 3\boxed 3\boxed 3\boxed 1\boxed 1\boxed 1 \;\;\boxed 3\boxed 3\boxed 2\boxed 2\boxed 1\boxed 1\;\;\boxed 3\boxed 2\boxed 2\boxed 2\boxed 2\boxed 1$
$\boxed 2\boxed 2 \boxed 2 \boxed 2 \boxed 2 \boxed 2$           
and we are interested only in the last pattern.
Now the number of ways for placing $12$ distinguishable balls in $6$ bins, the bins being indistinguishable  except by occupancy, the multinomial coefficient has to be divided by ${p! q!..}$ where $p$ bins each have $p_n$ balls, $q$ bins each have $q_n$ balls, etc, e.g.  for $\boxed 4\boxed 2\boxed 2\boxed 2\boxed 1\boxed 1$, there are $\binom{12}{4,2,2,2,1,1}/(3!2!)$ ways.
$S_2(12,6)= 1,323,652$ From this, if we subtract number of ways for $10$ unwanted patterns, we get  
$1323652 - \left[\binom{12}{7,1,1,1,1,1}/ 5!+\binom{12}{6,2,1,1,1,1,}/4! ...+\binom{12}{5,2,2,1,1,1}/(2!3!) + ... \binom{12}{3,2,2,2,2,1}/4!\right] = 10,395$
$10,395$ is the number of ways to place$2$ each of $12$ distinguishable balls in $6$ indistinguishable bins,
So $10,395\times 6! = 7,484,400$, the desired answer.
But this looks like trying to catch your nose by bringing your hand from behind your neck !
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