<s>[INST] In how many ways can $6$ red and $6$ blue balls be distributed among $10$ persons such that every person gets at least one ball? In how many ways can $6$ red and $6$ blue balls be distributed among $10$ persons such that every person gets at least one ball? (Balls of the same colour are identical).
My Attempt
There appear to be too many  cases which we need to consider. Not able to find correct approach.
 [/INST] Here is a solution by way of the Principle of Inclusion / Exclusion (PIE).
Suppose we ignore, for now, the restriction that each person must get at least one ball.  Then by a bars-and-stars argument, there are $\binom{10+6-1}{6}$ ways to distribute the six red balls among the ten people, and similarly for the six blue balls.  Since we can distribute the red and blue balls independently, the number of ways to distribute both the red and blue balls to the people is $N = \binom{10+6-1}{6}^2$.
Now to deal with the restriction that every person must get at least one ball.  Say a distribution of the balls has "Property $i$" if the $i$th person receives no balls, for $i=1,2,3,\dots,10$, and let $S_j$ be the number of distributions with $j$ of the properties, for $j=1,2,3,\dots,10$.  Then
$$\begin{align}
S_1 &= \binom{10}{1} \binom{9+6-1}{6}^2 \\
S_2 &= \binom{10}{2} \binom{8+6-1}{6}^2 \\
S_3 &= \binom{10}{3} \binom{7+6-1}{6}^2 \\
&\dots \\
S_9 &= \binom{10}{9} \binom{1+6-1}{6}^2
\end{align}$$
By PIE, the number of distributions with none of the properties, i.e. the number of distributions in which every person gets at least one ball, is
$$N_0 = N - S_1 + S_2 - S_3 + \dots - S_9 =26,250$$
</s>