The second bottle might have four choices if the one used for the first was the one intended for the second bottle. You are looking at the number of derangements. For $n$ items it is the closest natural to $\frac {n!}e$
The second bottle might have four choices if the one used for the first was the one intended for the second bottle. You are looking at the number of derangements. For $n$ items it is the closest natural to $\frac {n!}e$