Given 26 balls, 8 of them red and 18 of them blue, and 3 distinct, labeled groups to sort them in, how many different combinations can we make that satisfy the following conditions:
- All balls are used
- Each group has at least one of each color
- There are strictly more blue than red balls in each group
So I think I've simplified the problem a bit by ensuring some of the conditions are satisfied by reducing the number of balls to work with to 5 reds and 12 blues (since there has to be at least 1 red and 2 blues in each group). The part that's tripping me up is how I can count all the combinations while ensuring there are more blues than reds.
Remove the five remaining red balls and five blue balls and substitute purple balls for them. Now use stars and bars twice to ascertain that the five purple balls and seven blue balls can be distributed in $\binom{7}{5} \times \binom{9}{7}=756$ ways.
These are in $1$-$1$ correspondence with acceptable distributions of red and blue balls, as you can see by splitting each purple ball into a red ball and a blue ball.