Number of circular placements of $n$ identical letters such that no two letters are adjacent. Suppose I have to place $3$ identical letters on a circular table which has $7$ slots in such a way that no two letters are in consecutive slots. In how many ways can I do this?
Can this be generalized into $n$ identical letters on a circular table with m slots?
 A: Whenever you have a circular table, it's easiest to fix the position of one thing that you're placing, then place the rest relative to that position.
So, let's say your letters are $\color{red}{\operatorname{A}}$, $\color{blue}{\operatorname{B}}$, and $\color{green}{\operatorname{C}}$.   Let's fix $\color{red}{\operatorname{A}}$ in position $1$.  There are $6$ remaining positions.
$$\begin{array}{|ccccccc|} 
1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline
\color{red}{\operatorname{A}} & \times & \circ & \circ & \circ & \circ & \times \\ 
\end{array}$$
Note that $1$ is adjacent to $2$ and $7$.  So, neither $\color{blue}{\operatorname{B}}$ nor $\color{green}{\operatorname{C}}$ can go there.  If we place $\color{blue}{\operatorname{B}}$ in position $3$ or position $6$, we see that there are two places available for $\color{green}{\operatorname{C}}$.
$$\begin{array}{|ccccccc|} 
1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline
\color{red}{\operatorname{A}} & \times & \color{blue}{\operatorname{B}} & \times & \circ & \circ & \times \\ \hline \color{red}{\operatorname{A}} & \times & \color{blue}{\operatorname{B}} & \times & \color{green}{\operatorname{C}} & \times & \times \\\color{red}{\operatorname{A}} & \times & \color{blue}{\operatorname{B}} & \times & \times & \color{green}{\operatorname{C}} & \times \\
\end{array}\hspace{30pt}\begin{array}{|ccccccc|} 
1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline
\color{red}{\operatorname{A}} & \times & \circ & \circ & \times & \color{blue}{\operatorname{B}}  & \times \\ \hline
\color{red}{\operatorname{A}} & \times &  \color{green}{\operatorname{C}} & \times & \times & \color{blue}{\operatorname{B}}  & \times \\ \color{red}{\operatorname{A}} & \times & \times &  \color{green}{\operatorname{C}}& \times & \color{blue}{\operatorname{B}}  & \times \\
\end{array}$$
On the other hand, if $\color{blue}{\operatorname{B}}$ goes into positions $4$ or $5$, then there is only $1$ position left for $\color{green}{\operatorname{C}}$.
$$\begin{array}{|ccccccc|} 
1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline
\color{red}{\operatorname{A}} & \times & \times & \color{blue}{\operatorname{B}} & \times & \circ & \times \\ \hline
\color{red}{\operatorname{A}} & \times & \times & \color{blue}{\operatorname{B}} & \times & \color{green}{\operatorname{C}} & \times \\ 
\end{array}\hspace{30pt}\begin{array}{|ccccccc|} 
1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline
\color{red}{\operatorname{A}} & \times & \circ & \times & \color{blue}{\operatorname{B}} & \times & \times \\  \hline
\color{red}{\operatorname{A}} & \times & \color{green}{\operatorname{C}}& \times & \color{blue}{\operatorname{B}} & \times & \times \\ 
\end{array}$$
In each of these final positions, $\color{blue}{\operatorname{B}}$ and $\color{green}{\operatorname{C}}$ can be permuted (i.e. $\color{green}{\operatorname{C}}$ placed before $\color{blue}{\operatorname{B}}$), but doing so only moves us to a solution we already obtained.  (This also holds for the general case: the order of placement does not matter.  Why?) Thus we are left with $$2\times 2 + 2\times 1=6\text{ ways.}$$
Can you extend this counting technique to higher numbers of slots?
