Counting

X-Ploration into Counting

By Frater Elijah

1 Cut some square tiles from cardboard. Divide the top side of each tile into four congruent smaller squares and color them red, blue, green, and yellow, all squares of a tile with different colors.

a) What is the maximum number of such tiles that can be distinguished by arrangement of colors? (What does it mean to say that two tiles cannot be distinguished?)

Well we must first decide, 'what does it mean to say that two tiles cannot be distinguished'? This shall mean that two tiles are the same. So now, we are looking at the number of possible variances of tiles without repetition. The below is an example of two identical tiles, the only difference being with regard to rotation (90^o counterclockwise):

Now let us go about doing this. We first fix the upper left corner R. Then we vary the rest. We arrive at the following:

Now it can be easily seen that by fixing the upper left hand corner for the other colors will result in a repeated sequence of colors (in either the clockwise or counter clockwise direction). The following table lists all possible sequences (6 total):

Note: We are using sequences now to represent a particular cube formation in a clockwise direction.

# of different patterns	Precise pattern fixing the top left corner
1	RYBG	YBGR	BGRY	GRYB
2	RYGB	YGBR	BRYG	GBRY
3	RGYB	YBRG	BRGY	GYBR
4	RGBY	YRGB	BYRG	GBYR
5	RBYG	YGRB	BYGR	GRBY
6	RBGY	YRBG	BGYR	GYRB
All of these are the same
Another way to easily see this is to just continue repeating the pattern and start at any color to
find the start of the four sequences.
ex pattern 1: RYBGRYBGRYBGRYBGRYBGRYBGRYBG -If we start at say any Blue we can see the four
pattern of BGRY of which RYBG is equivalent to.

b) Extend the above problem for rectangles. An interesting observation of the above is that we can think of this as four different objects (4 colors), we fix one and then vary the rest, so for a 2 x 2 square we have (4-1)! = 3(2) = 6 different possibilities. Now when you say, extend the above case to rectangles there are many different ways to interpret this. I follow the following interpretation: a rectangle of size m x n of mn colors (each tiles color is distinct).

This case is like part (a). We have a total of mn cubes each of a different color. We fix one, and then their are (mn - 1)! different ways of computing the rest.

Fix the first cell, that leaves (mn -1) ways to choose the next cell different, then (mn - 2) ways to choose the next cell different and so on, to get (mn - 1)!

Note: Multiplication follows from the fundamental principle of counting.

Another interpretation of this problem is to arrange the four colors into sequences of four, then eliminate any duplicates under the operation of reflecting about a vertical line (180^o rotation). All possibilities are listed in Appendix A (24 results). Appendix B is listed with deletions (12 results). Note: This is the interpretation I believe you wanted.

Appendix A

(Possible variations of the four sequence)

RYBG	YBGR	BGRY	GRYB
RYGB	YGBR	BRYG	GBRY
RGYB	YBRG	BRGY	GYBR
RGBY	YRGB	BYRG	GBYR
RBYG	YGRB	BYGR	GRBY
RBGY	YRBG	BGYR	GYRB

Appendix B

(Possibilities with deletions of repeated sequences under 180^o rotation)

~~RYBG~~	~~YBGR~~	BGRY	GRYB
~~RYGB~~	~~YGBR~~	BRYG	GBRY
~~RGYB~~	~~YBRG~~	BRGY	GYBR
~~RGBY~~	~~YRGB~~	BYRG	GBYR
~~RBYG~~	~~YGRB~~	BYGR	GRBY
~~RBGY~~	~~YRBG~~	BGYR	GYRB

Does it make a difference if both sides of the tiles are colored?

Consider the following cases:

c) The coloring is such that the four squares of each tile are the same color on both sides.

This is simple. Think of it this way. We are defining one side of the square, which we have gives us six possibilities (as above). Now, the other side is static, that is, it is not changing (it is the same as the front). So how many possible variances will we have...? Well, six. Exactly the same ones as the front side. So, in this case it does not make a difference.

d) Both sides are independently colored R, B, G, Y. This is more interesting, as we want to formulate this without repetition. This is how I did it:

I first arranged the possible sequences with red fixed in the upper left hand corner in column 1 (this represents the front). Now in a horizontal row across the top I represented the possible arrangements for the back (there are six also). In a matrix form I then assigned coordinates to each possible combination in the form (a,b), where a is the row entry and b is the column entry (as per a standard matrix). Now we arrive at the following table:

	I	II	III	IV	V	VI
1	(1, I)	(1, II)	(1, III)	(1, IV)	(1, V)	(1, VI)
2	(2, I)	(2, II)	(2, III)	(2, IV)	(2, V)	(2, VI)
3	(3, I)	(3, II)	(3, III)	(3, IV)	(3, V)	(3, VI)
4	(4, I)	(4, II)	(4, III)	(4, IV)	(4, V)	(4, VI)
5	(5, I)	(5, II)	(5, III)	(5, IV)	(5, V)	(5, VI)
6	(6, I)	(6, II)	(6, III)	(6, IV)	(6, V)	(6, VI)

Now the entry (3, II) = (2, III), so we do not count this. As we can see all entries above the diagonal are repetitions of those below the diagonal. Summing these we arrive at (6 + 5 + 4 + 3 + 2 + 1) different arrangements of colors with red fixed in the upper left hand corner (the bold faced entries). We can trace this logic to the four other colors so we get 4 (6+5+4+3+2+1). Now how can we be sure that there are no repetitions? Remember that we fixed red on the front of the cube. If we fix another color in that same position then any arrangement involving that new color is guaranteed to be different from it's preceding color. Awesome! So we have 4 (21) = 84 possibilities to answer this question.

e) The below patterns can be folded to form a cube. Try to arrange the six tiles in such a way that at each corner of the cube adjacent colors are identical.