How many ways are there to put 6 balls in 3 boxes if three balls are indistinguishably white, three are indistinguishably black, and the boxes are distinguishable?

Let the boxes be Box 1, Box 2, Box 3. 

consider the 3 white balls. They can be all of them in one box:

(3, 0, 0) (3 in Box 1, 0 in box 2 and 0 in box 0)
(0, 3, 0)
(0, 0, 3)

We can have 2 in one box, and 1 in one of the remaining boxes:

(2, 0, 1)
(2, 1, 0)
(0, 2, 1)
(1, 2, 0)
(0, 1, 2)
(1, 0, 2)

and there is only one way: (1, 1, 1) to place one white ball in each box

In total there are: 3+6+1=10 ways to place the white balls. Similarly there are 10 ways to place the black ones.

Since every placement of the white balls can be combined with any placement of the black balls, there are 10*10=100 ways to place the 3white balls and the 3 black bals in the boxes.

100

Iwould say they are complementary because that’s the division of a 90 degree angle and it’s split between the middle of them