Distributing (not necessarily all) $6$ identical white and $8$ identical black balls to $3$ children

Solution 1:

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 $10268$