$9$ beads on $3$ by $3$ grid such that two of the beads are not adjacent How many ways are there to put $9$ differently colored beads on a $3\times3$ grid if the purple bead and the green bead cannot be adjacent (either horizontally, vertically, or diagonally), and rotations and reflections of the grid are considered the same?
I do not see a clear application of Burnside's Lemma because of the restriction that the beads can not be adjacent. I think that maybe casework or something would be good, but I can not do it due the number of cases. Are there any smart and slick solutions?
I tried making some progress on this question:
If the grid is denoted as follows:
\begin{array}{c *{10}{@{\;}c}}
&1& &2& &3&\\
&4& &5& &6&\\
&7& &8& &9&\\
\end{array}
Sorry for the bad alignment that is about to follow. Then the possible configurations are (please correct me if I am wrong)
\begin{array}{c *{10}{@{\;}c}}
&P& &2& &G&\\
&4& &5& &6&\\
&7& &8& &9&\\
\end{array}
\begin{array}{c *{10}{@{\;}c}}
&P& &2& &3&\\
&4& &5& &6&\\
&7& &8& &G&\\
\end{array}
\begin{array}{c *{10}{@{\;}c}}
&P& &2& &3&\\
&4& &5& &6&\\
&G& &8& &9&\\
\end{array}
\begin{array}{c *{10}{@{\;}c}}
&1& &P& &3&\\
&4& &5& &6&\\
&7& &G& &9&\\
\end{array}
\begin{array}{c *{10}{@{\;}c}}
&1& &P& &3&\\
&4& &5& &6&\\
&G& &8& &9&\\
\end{array}
\begin{array}{c *{10}{@{\;}c}}
&1& &G& &3&\\
&4& &5& &6&\\
&P& &8& &9&\\
\end{array}
I am stuck here. Cases $1$ and $3$ are identical, case $5$ and $6$ are identical, and cases $2$ and cases $4$ are separate (unless they are equal)? Can someone help me apply Burnside's Lemma?!
 A: Here is an alternate way / fuller explanation of my comment to the main thread.
I will use the term position to refer to one of the $9!$ possible arrangements of beads onto $9$ squares.
You want to count equivalent classes, where two positions belong to the same class if they are reflections/rotations of each other.  This problem is very simple because every class has $8$ positions, and more importantly, either all $8$ are valid or all $8$ are invalid.  So you can ignore all thinking about symmetry, and just count valid positions (i.e. treating reflections/rotations as distinct) and then divide by $8$.  This is what I meant by "Burnside's Lemma is not needed."

*

*If purple bead is at $1$, green can be at $3,6,7,8,9$.  So there are $5 \times 7!$ such positions.


*If purple bead is at $2$, green can be at $7,8,9$, so $3 \times 7!$ such positions.


*If purple bead is at $3$, green can be at $1,4,7,8,9$, so $5 \times 7!$ such positions.


*Etc.
In total there are $(5+3 + 5 + 3 + 0 + 3 + 5 + 3 + 5) \times 7! = 32 \times 7!$ valid positions.  Divide by $8$ and you get $4 \times 7!$ equivalent classes.
If you want to use symmetry-based thinking, the answer by Alex Ravsky is the way to go, but for this particular problem, that way is actually a bit more subtle and error prone.
A: Neither the purple nor the green bead can be put at the center of the grid. It follows that any placement of the beads can be transformed by reflections and rotations to exactly one of the following five groups $P_{ij}$ of placements, where the purple bead is placed at $i$ and the green bead is placed at $j$: $P_{13}$, $P_{16}$, $P_{19}$, $P_{27}$, and $P_{28}$. Each of groups $P_{13}$, $P_{16}$, and $P_{26}$ consists of $7!$ placements, because the placement of beads different from green and purple can be arbitrary.
Each of groups $P_{19}$ and $P_{28}$ consists of $7!/2$ placements, because the placement of beads different from green and purple splits into subgroups consisting of two symmetric placements (with the same positions of the purple and the green bead).  So there are $3\cdot 7!+2\cdot 7!/2=4\cdot 7!=20160$ unique bead  placements in total.
