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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.
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 $$ \sum_{r=4}^6 \binom{10}{r}\cdot\binom{5}{r-1}\cdot\binom{r + 5}{9} $$
There are the following cases.
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.
