Sara has $6$ flower pots, each having a unique flower. Pots are arranged in an arbitrary sequence in a row. Sara rearranges the sequence each day but no two pots should be arranged adjacent to each other, if they have already been adjacent to each other. For how many days can Sara keep rearranging the flowers (i.e. how many arrangements satisfying the rule above are possible)?
[Math] flower pot puzzle
combinatoricspuzzle
Related Solutions
As a simpler example, consider the case of 3 calls and 2 days. Your method would calculate the probability of getting a call each day as $$\frac{|\{(1,2),(2,1)\}|} {|\{(0,3),(1,2),(2,1),(3,0)\}|} = \frac12.$$
This assigns equal probability to all ways of splitting three indistinguishable calls into two days. However, this is not the probability distribution the question implies. If each call is independently and uniformly distributed among the two days, the tuples $(0,3)$ and $(3,0)$ have probability $1/8$ each (since it means that all three calls went to the same day, so $(1/2)^3$), while $(1,2)$ and $(2,1)$ three probability $\binom31(1/2)^3 =3/8$ each.
To summarise, uniform distribution for each call is not the same as uniform distribution over compositions (of $12$ calls into $7$ parts).
There are $\binom{6}{3} = 20$ sequences of three $A$'s and three $B$'s. Consider the ten sequences of three $A$'s and three $B$'s that begin with an $A$.
$\color{red}{AAABBB}$
$\color{green}{AABABB}$
$\color{green}{AABBAB}$
$\color{blue}{AABBBA}$
$\color{green}{ABAABB}$
$\color{cyan}{ABABAB}$
$\color{magenta}{ABABBA}$
$\color{green}{ABBAAB}$
$\color{magenta}{ABBABA}$
$\color{blue}{ABBBAA}$
No matter how we place $C$'s in the sequence $\color{red}{AAABBB}$, at least two consecutive letters will be the same.
There is only one way to place the $C$'s in the sequences $\color{blue}{AABBBA}$ and $\color{blue}{ABBBAA}$ since we are forced to place a $C$ between each pair of consecutive $B$'s and the pair of consecutive $A$'s.
The number of ways we can fill three of the seven spaces (the beginning, the end, and the five spaces between consecutive letters) in the sequence $\color{cyan}{ABABAB}$ is $\binom{7}{3}$.
We must place a $C$ between the pair of consecutive $B$'s in the sequences $\color{magenta}{ABABBA}$ and $\color{magenta}{ABBABA}$, which leaves us $\binom{6}{2}$ ways to insert the two remaining $C$'s in the six remaining spaces.
We must place one $C$ between the pair of consecutive $A$'s and another $C$ between the pair of consecutive $B$'s in the sequences $\color{green}{AABABB}$, $\color{green}{AABBAB}$, $\color{green}{ABAABB}$, $\color{green}{ABBAAB}$, leaving five spaces in which to place the remaining $C$.
Hence, the number of sequences of three $A$'s, three $B$'s, and three $C$'s that begin with $A$ that do not contain consecutive letters that are the same is $$1 \cdot 0 + 2 \cdot 1 + 2 \cdot \binom{6}{2} + 4 \cdot \binom{5}{1} + 1 \cdot \binom{7}{3} = 0 + 2 + 30 + 20 + 35 = 87$$ By symmetry, there are also $87$ such sequences that begin with a $B$. Hence, the number of sequences of three $A$'s, three $B$'s, and three $C$'s that do not contain consecutive letters that are the same is $2 \cdot 87 = 174$.
Best Answer
It seems the following.
If I understood you right then we have ${6 \choose 2}=15$ different pairs of pots. At each days Sara realizes 5 of these pairs as ajacent. Since all these pairs should be different, the number of days is at most $15/5=3$ . The following example describes the admissible list of arrangements for 3 days.