Number of ways to form groups of balls Suppose I have $12$ balls, $6$ are white and $6$ are black. I want to make four groups of balls. Each group contains three balls. How many arrangements are possible? Assuming all the same colored balls are identical.  
Thoughts: If all the balls would have been different the problem would have become simpler. I can arrange the balls in $12!$ ways and can argue that for each possible arrangement I can pick 3 balls at a time from the arranged sequence and form a group. I think a similar approach should work in identical balls case too. Giving the total number of ways as $\dfrac{12!}{6!6!}$. I don't know the answer to this problem. It will be really helpful if we can verify the answer with different methods.
 A: We have $6$ white balls (and therefore $6$ black balls) to place into $4$ bags. Each bag can contain $0,1,2$ or $3$ white balls (and correspondingly has $3-k$ black balls).
After each ball has been placed, we have a representative of the coefficient of $x^6$ in $(1+x+x^2+x^3)^4$.
So finding the coefficient of $x^6$ in $(1+x+x^2+x^3)^4$ gives the answer.
This can be calculated as:
$$[x^6]\left(\frac{1-x^4}{1-x}\right)^4$$
$$=[x^6]\frac{1-4x^4+\dots}{(1-x)^4}$$
$$=[x^6]\left(1-4x^4+\dots\right)\left(\sum_\limits{k=3}^\infty \binom{k}{3} x^{k-3}\right)$$
$$=\binom{9}{3}-4\binom{5}{3}$$
$$=84-4\cdot10$$
$$=44$$
A: Consider $4$ bins as groups. This problem is equivalent to calculating the number of partitions of $6$ comprising no more than $4$ numbers, with each group having 3 balls, after which the other- colored balls will fill the gap in each bin. You can manually count the partitions:  
$3,0,0,3$,
$1,2,0,3$,
$1,1,1,3$:
$1,2,1,2$
$2,2,2,0$
That’s $5$ ways. Note that this is assuming the groups are not distinguishable.
