Counting equivalence classes of $3\times3$ squares with $3$ cells labeled $a$, $b$, $c$

Question

Let's say we have a $3×3$ square where $3$ of the cells are labeled $a$, $b$, $c$ and the rest are blank. Two such squares are considered "equivalent" if one square can be obtained from another square by 1) rotation on 90, 180, and 270 degrees, 2) reflection (through the horizontal, vertical, or either diagonal axis).

I need to find equivalence classes of squares (maybe groups or patterns?).

My attempt is: 1) put the $a$, $b$ and $c$ in the 1-st row: $A=\left(% \begin{array}{ccc} a & b & c \\ .. & .. & .. \\ .. & .. & .. \\ \end{array} \right)$, then we can rotate the square $A$ on 90, 180 and 270 degrees: $A_{90}=\left(% \begin{array}{ccc} .. & .. & a \\ .. & .. & b\\ .. & .. & c \\ \end{array} \right)$, $A_{180}=\left(% \begin{array}{ccc} .. & .. & .. \\ .. & .. & ..\\ c & b & a \\ \end{array} \right)$, $A_{270}=\left(% \begin{array}{ccc} c & .. & .. \\ b & .. & ..\\ a & .. & .. \\ \end{array}. \right)$.

Four square $A$, $A_{90}$, $A_{180}$ and $A_{270}$ are equvalent. This is the first equivalence class.

2) put the $a$, $b$ and $c$ in the main diagonal: $$A=\left(% \begin{array}{ccc} a & .. & .. \\ .. & b & .. \\ .. & .. & c \\ \end{array}% \right) $$ and rotate on 90 degree $$A_{90}=\left(% \begin{array}{ccc} .. & .. & a \\ .. & b & .. \\ c & .. & .. \\ \end{array}% \right) $$ Two square $A$ and $A_{90}$ are equvalent. This is the second equivalence class.

Edit 2. Here I have found the 16 patterns.

Question. How many equivalence classes for three elements in a square are there?

Answer 1

It is unclear whether you want to count

1. Placements, or

2. Labelings.

up to symmetry. The two grids below have the same placement, but different labelings. The number of labelings without symmetry is $9\times 8\times 7$. The number of placements is $(9\times 8\times 7)/(3!)$.

a b c     a c b
. . .     . . .
. . .     . . .

---

Placements

There are $8$ symmetries of array:

• The identity symmetry.

• Three rotations, by $90,180$ and $270$ degrees.

• Four reflections, through the horizontal, vertical, or either diagonal axis.

In order to count the number of placements up to symmetry, we add up, for each symmetry, the number of placements which invariant under that symmetry, then divide by $8$.

• Each of the $\binom{9}3$ placements is invariant under the identity.

• There are no placements with three numbers which are invariant under $90^\circ$ rotation.

• There are $4 $ placements which are invariant under $180^\circ$ rotation. These are placements which have a three in a row through the center.

• For the horizontal reflection, there are $10$ placements with that symmetry. There are two cases; either two of the numbers are symmetric about the vertical axis, going in spots A, B or C, or they are all on the vertical axis. In the first case, there are $3$ choices for A, B or C, ,and three choices for where the last number goes. The second case has one option, for $3\cdot 3+1=10$ total.

A 1 A
B 2 B
C 3 C

• The same answer is true for the other three reflections.

Therefore, the number of placements up to symmetry is

$$ \frac18\left(\binom{9}3+4+4\cdot 10\right)=\frac18(84+44)=16 $$

Labelings

I leave the details to the reader, but the answer is

$$ \frac18(9\cdot 8\cdot 7+3\cdot 0+4\cdot (3\cdot 2\cdot 1))=66 $$