12 balls in a box to distribute by 2 bags. There is a box with 12 balls inside, 4 green, 5 blue and 3 black.
We want to take out 4 balls for each one of 2 bags (bag A and bag B), and we take note of the number of balls of each color that the bags have.
How many ways are there to distribute the balls?

We could start by figuring out the number of ways that $4$ balls do NOT go in either bag (that would also determine the $8$ that do).
The possibilities are:


*

*4 green $\displaystyle \Rightarrow \dbinom{4}{4} = 1$

*3 green, 1 blue $\displaystyle \Rightarrow \dbinom{4}{3} \dbinom{5}{1} = 4\cdot5 = 20$

*3 green, 1 black $\displaystyle \Rightarrow \dbinom{4}{3} \dbinom{3}{1} = 4\cdot3 = 12$

*2 green, 2 blue $\displaystyle \Rightarrow \dbinom{4}{2} \dbinom{5}{2} = 6\cdot10 = 60$

*2 green, 1 blue, 1 black $\displaystyle \Rightarrow \dbinom{4}{2} \dbinom{5}{1} \dbinom{3}{1} = 6\cdot5\cdot3 = 90$

*2 green, 2 black $\displaystyle \Rightarrow \dbinom{4}{2} \dbinom{3}{2} = 6\cdot3 = 18$

*1 green, 3 blue $\displaystyle \Rightarrow \dbinom{4}{3} \dbinom{5}{3} = 4\cdot10 = 40$

*1 green, 2 blue, 1 black $\displaystyle \Rightarrow \dbinom{4}{3} \dbinom{5}{2} \dbinom{3}{1} = 4\cdot10\cdot3 = 120$

*1 green, 1 blue, 2 black $\displaystyle \Rightarrow \dbinom{4}{3} \dbinom{5}{1} \dbinom{3}{2} = 4\cdot5\cdot3 = 60$

*1 green, 3 black $\displaystyle \Rightarrow \dbinom{4}{3} \dbinom{3}{3} = 4\cdot1 = 4$


There are also different ways to place $4$ balls into bag $A$ (which would in turn decide which $4$ go into bag $B$). I leave that for you to do.
