Here's another way for 1 using Complementary Counting:
There are \({8 \choose 5} = 56\) ways to order them.
If only 2 white chairs are together, there are 7 positions they can occupy (1 and 2, 2 and 3, 3 and 4, and so on)
If the white chairs occupy spots (1 and 2) or (7 and 8), there are 5 options on how to order the remaining chairs, making for 10 ways (the white chair can't be directly next to the white chairs).
For the remaining 5 spots, there are only 4 ways to put the remaining white chair, making for 20 ways.
This means that there are \(5 \times 4 + 5 + 5 = 30\) ways if only 2 white chairs are next to each other.
Next, there are only 6 ways for all 3 white chairs to be next to each other (1,2, and 3; 2, 3, and 4; and so on).
This makes for \(56 - 30 - 6 = \color{brown}\boxed{20}\) ways.