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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.
A

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.

jiten
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    I don't follow what you don't understand. A coloring of type $(4,0,0)$ would be a coloring where there is a color such that all four of the objects are that color and the remaining colors each have zero objects of that color. – JMoravitz Feb 28 '20 at 15:04
  • @JMoravitz Do you mean that all $4$ fields if colored by the same color, then denoted by $(4,0,0)$. – jiten Feb 28 '20 at 15:06
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    Yes... for example the coloring where all four of the squares are colored red. Also the coloring where all four of the squares are colored blue... there are three total colorings of this type (all red, all blue, or all yellow). Compare this to the colorings of type $(3,1,0)$ where there is a color such that three of the squares are that color and there is one square colored a different color while the final color has zero squares colored that color... for example the arrangement where you have three squares colored red and one square colored blue. There are as mentioned six such colorings – JMoravitz Feb 28 '20 at 15:08
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    These triples represent the multiplicities of each colour. – Donald Splutterwit Feb 28 '20 at 15:10
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    Note the subtlety of the $(2,2,0)$ case. You might be in a situation like $\begin{bmatrix} 1&1\0&0\end{bmatrix}$ where the colors might be adjacent to others of its color, or you might instead be in a situation like $\begin{bmatrix}1&0\0&1\end{bmatrix}$ where they are not. – JMoravitz Feb 28 '20 at 15:11
  • @JMoravitz But, still confused as to how the $(2,2,0)$ case has only $6$ possible configurations for four squares labelled $a,b,c,d$. I take fields' coloring in your matrix from top right in cck direction: 1. RRBB, 2. RRYY, 3. RBRY, 4. RYRB, 5. BBRR, 6. BBYY, and more should follow. So, am definitely wrong. – jiten Feb 28 '20 at 15:17
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    You say RBRY. That would not be in the $(2,2,0)$ case but rather in the $(2,1,1)$ case. The possibilities are in alphabetical order (earliest representation of each used): BBRR, BBYY, BRBR, BYBY, RRYY, RYRY. Note for instance that since rotations are considered identical, I only included BBRR once rather than also including BRRB, RRBB, and RBBR since these are all considered identical to BBRR. Note that in each of these, we have two colors, each of which used twice. RYRB for instance has three colors used, one used twice, another used once, and yet another once – JMoravitz Feb 28 '20 at 15:20
  • @JMoravitz Sorry, for mixing up the two cases of $(2,2,0)$ and $(2,1,1)$. For the $(2,1,1)$ case have $7$ cases as: $1. RRBY, 2. RRYB, 3. RBRY, 4. BBRY, 5. BBYR, 6. BRBY, 7. YYRB, 8. YYBR, 9.YRYB$. I think $RYRB$ cannot be included as by rotation it is the same as $RBRY$. Please vet my cases. – jiten Feb 28 '20 at 15:34
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    For the $(2,1,1)$ case, it will be easier not to write them in alphabetical order for me, but rather specifically in such a way as to have the pair first if it exists. We have the six where the pair is together: BBRY, BBYR (note that these two are different), RRBY, RRYB, YYBR, YYRB, and then the three where the pair is apart RBRY, BRBY, YBYR. As an aside, learning to count these without brute force listing them out is one of the things introductory combinatorics attempts to teach you. Brute force listing them out should only be used for intuition, ease, or last resort. – JMoravitz Feb 28 '20 at 15:38

2 Answers2

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$(4,0,0)$ 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 $6+3=9$.

Siong Thye Goh
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  • Thanks for your answer, request a python program for listing the possible combinations for the different cases of the problem, as say for the case of (2,1,1). – jiten Feb 28 '20 at 16:10
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    here. – Siong Thye Goh Feb 28 '20 at 16:39
  • Seems like need discuss how program works, in the last chatroom (https://chat.stackexchange.com/rooms/104996/discussion-between-siong-thye-goh-and-jiten). – jiten Feb 28 '20 at 17:00
  • Please help with a doubt in equivalent grid rotations, raised in my today's post at: https://math.stackexchange.com/q/3564301/424260. – jiten Feb 29 '20 at 15:09
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$(4,0,0)$ represents all the colorings in which four fields have same color. For a coloring with 2 yellow, 1 red and 1 blue fields, the coloring is of $(2,1,1)$ type.A coloring with 1 yellow,2 red, and 1 blue is also of the type $(2,1,1)$.

In case of $(4,0,0)$, all the four fields are of same color and they can be of red,yellow or blue.So three such combinations are possible. $(2,2,0)$ represents a coloring with just two colors and when a distinct coloring is rotated it should not give any other coloring. We can only do this in two ways. Example of(2,2,0) coloring

For three colors this is $$^3C_2 \times2 =6 $$ combinations

callculus42
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fruitbat
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