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?
 A: 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.
