Randomly distributing 2 Red, 3 Green, 4 Blue balls into 3 bags with each bag containing exactly 3 balls. What is the probability
- exactly one bag contains all colors?
- exactly two bags contain all colors?
- no bag contains all colors?
For second part: $2 \cdot 3 \cdot 4$ ways for bag 1, remaining $1 \cdot 2 \cdot 3$ for bag 2 and rest balls with $3!$ ways. Since any bag can have this combination we multiply by $3!$ and then divide everything by $9!$ to get the probability. Now for first part, I think we can use this same method but after giving balls to bag1 we can have rest balls with $6!$ ways. this will also include some cases from second part so we need to subtract them first using principle of inclusion and exclusion.