In the book titled: Combinatorics: A Problem-Based Approach, by: Pavle Mladenović; in section 2.7; there is a solved problem (2.7.11) for finding by direct counting the number of non-equivalent colorings.
There are given $4$ unit square parts (called fields) of a square of size $2*2$, that are to be colored with three colors – red (R), blue (B), yellow(Y).
The problem description is: two colorings are considered to be the same if the first of them can overlap the second by rotating the given square of size $2*2$. Find the number of non-equivalent colorings under this assumption.
The answer given by the book states:
We say that a coloring is of $(a, b, c)$-type, where $a\geq b\geq c$, if $a$ of the given four fields are colored the same color (red, blue, or yellow), and $b$ and $c$ of the fields are colored the other two colors. Possible types of colorings are
$(4, 0, 0), (3, 1, 0), (2, 2, 0)$, and $(2, 1, 1)$. It is easy to see that there are $3, 6, 6$, and $9$ colorings of these types, respectively. The total number of distinct colorings is $3 + 6 + 6 + 9 = 24$.
I have confusion in understanding the notations used – say, how there is value $0$ when $4$ is also there. I mean the integers $0,1, 2,3,4$ signify $5$ values; that I do not see in any category – number of fields, or number of colors. Seems, am lost in understanding of the approach taken by the author.
Best Answer
$(4,0,0)$ means all the colors are the same, hence we can choose one of the $3$ colors.
$(3,1,0)$ means we pick $3$ units to be of one color and $1$ unit to be of different color. We can pick two colors out of $3$ and we can swap them. $\binom32 \cdot 2=6$.
$(2,2,0)$ corresponds to two distinct pattern, of which they might be $\begin{bmatrix} 1 & 1 \\ 0 & 0\end{bmatrix}$ or $\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$, after choosing one of them, we get to choose $2$ colors out of $3$. That is $2 \cdot \binom32=6$
$(2,1,1)$ corresponds to two distinct pattern, of which they might be $\begin{bmatrix} A & A \\ B & C\end{bmatrix}$ or $\begin{bmatrix} A & B \\ C & A\end{bmatrix}$. If the first one is chosen, we get to choose $(A,B,C)$ in $3!=6$ options. If the second one is chosen, notice that $B$ and $C$ can be rotated to each other, hence we have $\frac{3!}2=3$ options. In total $6+3=9$.