How many ways are there to distribute 2 indistinguishable white and 4 indistinguishable black balls into 4 indistinguishable boxes?
How can we solve this?
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.
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.
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.
Once again: 2-0-0-0, 1-1-0-0, 1-0-1-0, 0-0-2-0, 0-0-1-1. 5 ways.
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.
Lastly: 2-0-0-0, 1-1-0-0. 2 ways.
In total, $4+7+5+7+2 = 25$ ways to distribute the balls.
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.
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.
To distribute 4 black balls to 4 boxes, you can do the following
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!