Number of ways to distribute five red balls and five blues balls into 3 distinct boxes with no empty boxes allowed

Question

Find the number of distributions of five red balls and five blues balls into 3 distinct boxes with no empty boxes allowed

so I know if I have just 5 identical balls and 3 distinct boxes, the answer would be $\binom{7}{2}$ but because i have another set of 5 balls, I'm unsure how to proceed with this question.

I believe there is a way to use the inclusion/exclusion principle. Set of red balls and set of blue balls and subtract the set that coincides?

Answer 1

Yes. PIE seems to be the way to go.

There are $\tbinom{3}{3}\tbinom{5+3-1}{3-1}^2 - \tbinom{3}{2}\tbinom{5+2-1}{2-1}^2+\tbinom{3}{1}\tbinom{5+1-1}{1-1}^2$ ways to do so.