3 Answers
3

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
$$N_0 = N - S_1 + S_2 - S_3 + \dots - S_9 =26,250$$

For each combination, let $r$ be the # of people that have at least one red ball. We have $4 \leq r \leq 6$. There are $\binom{10}{r}$ ways to select $r$ people and $\binom{6-1}{r-1}$ ways to distribute the $6$ red balls to the $r$ people selected so that each one has at least one. For the remaining $10 - r$ people, they must have at least $1$ blue ball. Therefore, there are $6 - (10 - r) = r - 4$ blue balls remaining and there are $\binom{(r-4) + 10-1}{10-1}$ ways to distribute the $r-4$ blue balls to the $10$ people. We conclude
$$
\sum_{r=4}^6 \binom{10}{r}\cdot\binom{5}{r-1}\cdot\binom{r + 5}{9}
$$

one person gets three balls of the same colour, everyone else gets one ball

one person gets three balls, two of which are the same colour, everyone else gets one

two people get two balls, all of the same colour

two people each get a red and a blue ball, everyone else gets one

one person gets two balls of the same colour, one person gets a red and a blue ball

one person gets two red balls, another gets two blue balls

As an example, case 6 results in $\binom{10}2\times\binom 84\times 2$ different ways, because there are $\binom{10}2$ ways to choose the two people who get two balls, $2$ ways to choose which of them gets blue balls and which red, and then $\binom 84$ ways to choose which of the remaining $8$ people get the $4$ remaining red balls. You can do the other cases similarly and add them up.