How many ways are there to distribute 2 indistinguishable white and 4 indistinguishable black balls into 4 indistinguishable boxes? How many ways are there to distribute 2 indistinguishable white and 4 indistinguishable black balls into 4 indistinguishable boxes?
How can we solve this? 
 A: There are 5 essentially different ways to distribute the black balls. In each case I'll count the essentially different ways of distributing the white balls.

4, 0, 0, 0

The possible distinguishable ways of distributing the white balls are: 2-0-0-0, 1-1-0-0, 0-2-0-0 and 0-1-1-0. So 4 is the number.

3, 1, 0, 0

Here we can do it like this: 2-0-0-0, 1-1-0-0, 1-0-1-0, 0-2-0-0, 0-1-1-0, 0-0-2-0, 0-0-1-1. So 7.

2, 2, 0, 0

Once again: 2-0-0-0, 1-1-0-0, 1-0-1-0, 0-0-2-0, 0-0-1-1. 5 ways.

2, 1, 1, 0

And again: 2-0-0-0, 1-1-0-0, 1-0-0-1, 0-2-0-0, 0-1-1-0, 0-1-0-1, 0-0-0-2. 7 ways.

1, 1, 1, 1

Lastly: 2-0-0-0, 1-1-0-0. 2 ways.
In total, $4+7+5+7+2 = 25$ ways to distribute the balls.
A: The numbers are small enough to allow a fairly brute-force approach. First break down the possibilities according to the number of non-empty boxes.


*

*We can have all six balls in one box; that’s $1$ distribution.

*If exactly two boxes are non-empty, the balls can be split $5$-$1$, $4$-$2$, or $3$-$3$. There are $2$ distinguishable $5$-$1$ splits, one with a singleton white ball, the other with a singleton black ball. There are $3$ distinguishable $4$-$2$ splits: the $2$ can be $2$ white balls, $2$ black balls, or one of each color. There are $2$ distinguishable $3$-$3$ splits, one with the $2$ white balls in the same box and one with them in different boxes. That’s a total of $7$ distributions.

*If exactly three boxes are non-empty, the balls can be split $4$-$1$-$1$, $3$-$2$-$1$, or $2$-$2$-$2$. There are $3$ distinguishable $4$-$1$-$1$ splits: one with two white singletons, one with two black singletons, and one with a singleton of each color. There are $5$ distinguishable $3$-$2$-$1$ splits: one with the two white balls as the pair, two with a pair of black balls, and two more in which the pair contains one ball of each color. There are only $2$ distinguishable $2$-$2$-$2$ splits, one with the white balls in the same box and one with them in different boxes. That’s a total of $10$ distributions.

*If all four of the boxes are non-empty, the balls can be split $3$-$1$-$1$-$1$ or $2$-$2$-$1$-$1$. Can you finish the counting from here?
A: Maybe you can treat this as two separate problems.
To distribute 2 white balls to 4 boxes, you can either put 2 balls in one box or put 1 ball in each of two boxes.
1.|WW||||
2.|W|W|||
To distribute 4 black balls to 4 boxes, you can do the following  
3.|BBBB||||
4.|BBB|B|||
5.|BB|BB|||
6.|BB|B|B||
7.|B|B|B|B|
Now how do you combine them?? To combining 1 and 3 gives you a total of 2 possibilities (put all the balls in the same box or separate them into different boxes).
To combine 2 and 3, you can put all the white balls in a box with a black ball or a box without a black ball, so 2 total ways.
To combine 1 and 4, you know you can put the white balls in a box with 3 black balls, 1 black ball ,or no black balls, so 3 total ways.
To combine 1 and 5, there are 2 total ways.
To combine 6 and 1, there are 3 total ways.
To combine 7 and 1, there is 1 way.
It gets a bit trickier now. To combine 2 and 4, you pair up each white ball with a box with at least 1 black ball. Otherwise, put only 1 white ball in a box without a black ball, and put the other white ball in either of the two black ball contained boxes. Lastly, you can put the two white balls each in a box without any black balls, giving you a total of 4 ways. 
To combine 2 and 5, follow the same pattern as right above, but there is one less way to combine the two since you can putting a single white ball in either of the black ball contained boxes is indistinguishable (so 3 total ways)
To combine 2 and 6, you can put the two whites with the (BB,B),(B,B),(B,x),or (BB,x), for 4 total ways.
Lastly, for 2 and 7, there is clearly 1 total way.
Add up all the numbers and get 25!
