Distributing (not necessarily all) $6$ identical white and $8$ identical black balls to $3$ children 
We have $6$ identical white balls and $8$ identical black balls. In how many ways can we distribute the balls to $3$ children so that each gets at least one? We don't necessarily have to distribute all $14$ balls we have.

My thoughts:
There are already similar questions and a generalized answer to one.
In my task, I thought of separating the problem into $2$ cases: when we distribute all $16$ balls to $3$ children and when we don't. For the latter, I thought I could bring an 'auxiliary' child that should also get at least one ball. So, let $A_i=\{\text{ distributions where }i^{\mathrm{th}}\text{ child gets no ball }\}.$ In the first case we're looking for $$\begin{aligned}\left|\bigcap_{i=1}^3 A_i^c\right|&=\left|\left(\bigcup_{i=1}^3A_i\right)^c\right|\\&=\binom{3+6-1}6\binom{3+8-1}8-\left|\bigcup_{i=1}^3 A_i\right| \quad (1)\end{aligned}$$
We can now compute $(1)$ analyzing when $i,1\le i\le 3$ children get no balls, that is $3-i$ of them do, which can be done in $\binom3k\binom{3-i+6-1}6\binom{3-i+8-1}8=\binom{3}i\binom{8-i}6\binom{10-i}6$ ways. Therefore $(1)$ equals $$\begin{aligned}\sum\limits_{i=0}^3(-1)^i \binom3i\binom{3-i+6-1}6\binom{3-i+8-1}8&=\sum\limits_{i=0}^3(-1)^i\binom{3}i\binom{8-i}6\binom{10-i}8\\&=28\cdot45-3\cdot7\cdot9+3\\&=1074\end{aligned}$$
Now, for the second case, analogously, $$\begin{aligned}\left|\bigcap_{i=1}^4A_i^c\right|&=\sum_{i=0}^4(-1)^i\binom4i\binom{9-i}6\binom{11-i}8\\&=84\cdot165-4\cdot28\cdot 45+6\cdot7\cdot 9-4\\&=9194\end{aligned}$$
so my answer is $\#=10268$
Could somebody check this solution?
 A: This is not as nice as your solution, but I get the same answer.
So each child has $3$ choices: receive only white balls, receive only black balls, or receive both. Represent these conditions as $-1,1,0$ respectively. If we have $x$ $-1$'s $y$ $1$'s and $z$ $0$'s, then we will first subtract give the $x+z$ children $1$ white ball and the $y+z$ children $1$ black ball. We are left with $6-x-z$ white balls and $8-y-z$ black balls to distribute accordingly.
The remaining $6-x-z$ white balls must go to the $x+z$ children who will receive white balls (or to no one). We can define an auxilary child who will receive a nonnegative amount of unused balls. We can then see that there are $\binom{6}{x+z}$ ways to distribute these by stars and bars. Similarly there are $\binom{8}{y+z}$ ways to distribute the black balls.
Because there must be a total of $3$ $-1$'s, $1$'s, and $0$'z we should sum over all $x+y+z=3$. However, there are $\binom{3}{x,y,z}$ ways to arrange these conditions (i.e. assign children which choice they get), so our answer is
$$\sum_{\substack{x+y+z=3\\x,y,z\geq 0}} \binom{3}{x,y,z}\binom{6}{x+z}\binom{8}{y+z}$$
We can change the bounds into
$$=\sum_{z=0}^3 \sum_{\substack{x+y=3-z\\x,y\geq 0}} \binom{3}{x,y,z}\binom{6}{x+z}\binom{8}{y+z}$$
$$=\sum_{z=0}^3 \sum_{y=0}^{3-z} \binom{3}{3-y-z,y,z}\binom{6}{3-y}\binom{8}{y+z}$$
$$=\sum_{z=0}^3 \binom{3}{z}\sum_{y=0}^{3-z} \binom{3-z}{y}\binom{6}{3-y}\binom{8}{y+z}$$
I'm not sure if this has any nicer form, but possibly exponential generating functions could be helpful. However, one can evaluate with a calculator that this sum is $10268$
