All wrong (Solution)

Total number of ways to arrange 8 balls? \(8!\)

Total number of ways that I can arrange the balls s.t. ball #1 is in box #1? \(7!\)

Total number of ways that I can arrange the balls s.t. ball #2 is in box #2? \(7!\)

Etc..

So the total number of ways we can arrange balls s.t. no ball is in the correct box?

\[8! - 8 \cdot 7!\]

But wait, what about any ordering where both ball #1 and ball #2 are in the correct place. We over substracted those orderings, once for when ball #1 is in the right spot and once for when ball #2 was in the right spot. So, for any pair of balls, let’s add back the number of ways that pair of balls can both be in the right spot:

\[8! - 8 \cdot 7! + {8 \choose 2} 6!\]

But wait, what about any ordering where balls #1, #2, and #3 are the the correct place? First, I subtracted them 3 times, then I added them back \({3 \choose 2} = 3\) times, so we have to subtrace one again. And we have to do that for every triple of balls.

\[8! - 8 \cdot 7! + {8 \choose 2} 6! - {8 \choose 3} 5!\]

You can see where this is going. Number of ways we can arrange the balls s.t. no ball is in the correct place:

\[\sum_{x=0}^8 (-1)^x {8 \choose x} (8-x)!\]

So the probability I randomly arrange the balls s.t. no ball is in the same numbered box?

\[\frac{\sum_{x=0}^8 (-1)^x {8 \choose x} (8-x)!}{8!}\]