Layers
I am carrying out an investigation to find out the different arrangements of cubes on a specified grid size. I will first start off with a two by three grid size which means there are six squares in the grid. On these six squares I will put five cubes. Each cube must fit exactly onto one square. During the course of my investigation I will display and describe my work and findings.
I first investigated how many different arrangements of the five cubes there were on a 2 by 3 grid. The 2 by 3 grid obviously has 6 squares; as one square always has to stay blank, the other five squares will be filled in. Because there are six squares and one square is always blank, there are six different variations.
For the second layer, there are only five squares, one of which has to stay blank. Because of this four squares can be filled, and this produces 5 different arrangements.
As you can see the total number of different combinations is 30. This can also be worked out by saying that there are five different arrangements on the second layer and there are six different arrangements on the first layer. So if you calculate 5*6 you get the answer 30.
There is a theory behind this to find out the number of different arrangements without drawing the layers:
"The number of squares filled in is always -1 of the number of arrangements and the number of possible empty squares."
To test this theory out I did a made another grid of 3 by 2. Although this is the same as the 2 by 3 grid, I wanted to make the increase in grid size steady. I got the same results as the 2 by 3 grid.
The next grid size that I used was a 3 by 3 grid.
In a 3 by 3 grid, there are nine squares. This means there are 9 possible empty squares and 8 possible filled in squares. This is proof that the number of filled in squares is one less than the number of possible empty squares.
Therefore there are 9 different arrangements of cubes. To further test this theory, I used a 3 by 4 grid. In a 3 by 4 grid there are 12 squares. Of this 12 squares are possible empty squares and 11 are possible filled in squares. This also supports the theory of the number of squares that are filled in is -1 of the number of arrangement and the number of possible empty squares.
On the second layer of a 3 by 3 grid, there will be eight squares. This means that there are eight possible empty squares and there are seven possible filled in squares. Even on the second layer the rule still works.
The reason why there are only eight square on the second layer when there were nine on the first layer is according to 'Rule 2: Each new layer is made with one cube less than the layer underneath it.'
Also on the 3 by 4 grid, there are 12 squares. On the first layer, there are 12 squares. This means that there are twelve possible empty squares and 11 possible filled in squares. On the second layer there are 11 possible empty in squares and 10 possible filled in squares. Here the theory holds true again.
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The reason why there are only eight square on the second layer when there were nine on the first layer is according to 'Rule 2: Each new layer is made with one cube less than the layer underneath it.'
Also on the 3 by 4 grid, there are 12 squares. On the first layer, there are 12 squares. This means that there are twelve possible empty squares and 11 possible filled in squares. On the second layer there are 11 possible empty in squares and 10 possible filled in squares. Here the theory holds true again.
It is also possible to know how many cubes will be missing and how many arrangements there are for any given layer. To determine how many arrangements there are, you have to know how many squares are on the specific layer.
If you have to find out the number of squares on the second layer; subtract 2 from the number of squares on the grid. For the 1st layer, subtract 1 from the number of squares on the grid.
To find the number of arrangements, count the number of squares on the layer. If you starting with the grid, the number of squares = the number of arrangements. To find the number of arrangements, the rule is "the number of squares on the grid/layer = the number of arrangements".
To prove this theory, I again used the 3 by 3 grid. On the first layer there are 9 squares on the grid. I believe there will be 9 different arrangements.
The reason behind this is that because there are nine squares, and each square is a possible empty square, there will be nine arrangements due to the fact that each square will have to be empty at one time or another.
Now that I have found that the two formulas hold true, I will have to produce formulae so as to put the formulae to work and that the answer can be produced without having to go into drawing so many squares and cubes.
If you let n be the number of squares on the grid, and the number of cubes on the first layer is n-1.
Let the number of variations be n.
For the second layer; let the number of cubes be n-2 and the number of variation n-1.
To test this out I used a 2 by 4 grid, in the grid there are 8 squares.
As you can see, if you take n to be the number of squares on the grid: here n=8 and the number of cubes is seven. This can be n-1. The number of arrangements is also eight. This means that the number of arrangements is n. As you can see, all the above corresponds to the formula and proves it right.
I will now have to show this onto the second layer.
Due to the complexity of doing the second layer for eight different arrangements, I will only show one arrangement. Another reason is that if it is right for one layer it is correct to surmise that the other layers will be the same.
As you can see, all the cubes shown are the second layer. There are six cubes on all the arrangements, this is n-2, n=8. Also, the number of arrangements is 7, this is n-1.
But this is not all; I have to make a formula for all the layers starting from the first layer up to the last layer to find the total number of arrangements.
As you can see, the above formulas are a bit complex. As I have said before, the number of arrangements on the first layer is n the same as the number of squares on the grid. But on the second layer it is n-1. That is when I realized that the two formulas are related.
I will show the third layer of one of the arrangements shown above. The reason is as above due to complexity.
Here you can see that from the bottom layer of eight arrangements, there are now, in the third layer, six arrangements. This can also be written as n-2.
From this so far, I can ascertain that it will work out as follows. So for the first layer the number of arrangements is n, for the second layer it is n-1, for the third layer it is n-2. Therefore it is possible to work out that for the fourth layer it will be n-3, for the fifth layer it is n-4, for the sixth layer it is n-5 and so on. But when we have reached, the last layer it is n-7. But if the grid size is larger than 2 by 4, it cannot work out to be n-7. So it makes sense to say that we can make the formula n-n+1.
In all the above formulas n means the number of squares on the grid. In the last layer, the formula n-n+1 stands for the number of squares on the grid subtracted by the same number and then add 1. The reason why you have to add 1 is that if you do n-n you will get the number 0. But if you add 1 you get the final number of 1.
As you can see, this is a rather complex way of working out the total number of arrangements. So I conducted some research and found out something called the factorial of a natural number. I also found that this can be worked out on a standard scientific calculator with a touch of a button. This button looks like: x! . In mathematics, the factorial of a natural number n is the product of all positive integers less than or equal to n.
In Part 3: "Investigate the relationship between the number of arrangements and the size of the grid when the number of empty squares on the first layer is greater than one", I will investigate the relationship between the umber of arrangements and the number of empty squares, when there are more than one empty square missing on the first layer.
As you can see I have drawn out all the different arrangements of a 2 by 3 grid which is six squares. On this grid there will be two empty squares which means that there will be four cubes that I can use. After drawing these out I have found that there are 15 arrangements. Now I have realised that in the beginning of my investigation I had one empty square on the grid and I got 6 different arrangements. Now I have two empty squares, and I have 15 different arrangements.
Because I have not found any relationship I will put two empty squares on a 2 by 4 grid which has 8 squares.
I did not draw the different arrangements of 2 by 4 grid because of its complexity. However I did draw it on a spare piece of paper in a 2d shape.
After drawing out all the different arrangements of a 2 by 4 grid with two empty squares, I have found 28 different arrangements. This means that my prediction was wrong.
That is why I will look for how many different arrangements there are in 2 by 5 grid with two empty squares. I will not draw out all the different arrangements because of its complexity. I will work it out on a spare sheet using 2d shapes. After this I will look for any kind of relationship between the number of arrangements and the size of the grid.
I have found out that a 2 by 5 grid has 48 different arrangements.
I will now attempt to find out a formula like the one that I first did, to find out how many different arrangements there can be with 2 squares missing on the first layer. However, on the second and third layers there will only be one square missing.
Through trial and error, which is an established mathematical method, I worked out that what the formula is.
If I take s to be the number of squares on the grid, then s-n is the number of squares subtracted by the number of the layer (n). The total number of layers that will be used has to be 2 less than the number of the grid size.
This is because, one the 1st section there is only 1 cube missing. So by the time you get down to 4 squares you are on the 2nd layer. If this is on the 3rd section which has 2 squares missing you will get down to 4 squares on the first layer.
As a result, in the first formula, the highest you can go up to is n-n+1, this mean that on the 2nd formula the highest you can go up to n-n+2.
The reason as to why the formula has to be divided by 2 is that if there was only one cube missing then there would be more different arrangements. So if there is only one cube missing, then that means the formula will have to be divided by 1. So it is logical to follow onto saying that because there are two cubes missing then the formula has to be divided by 2.
Another reason as to why this is is that when there is only 1 square missing then the total number of arrangements on the first and second layers is 30. When there are only 2 cubes missing, then the total number of arrangements on the first and second layers is 15. This is half of the total number of arrangements when the formula is divided by 1 and when there is only one cube missing.
The formula is: S(S-1)(S-2)(S-3)(S-4)(S-S+1)
To test this formula out, I saw that on the first layer of cubes, when there are only 2 cubes missing, the total number of arrangements is 15. Because there are only four cubes to work one, that means the scond layer will have four different arrangements. So 15*4 = 60.
Now, if you follow the formula, replacing the letters with numbers the formula turns out to be:
6(6-1)(6-2)
This can be calculated to give the answer: 60. This shows that the formula works. To further test the formula, you can see that it is apparent that, because of only one square being missing from the second layer onwards, you can see that the third layer will have 3 arangements. So this is 15*4*3 = 180. After I have incorporated this into the formula (6(6-1)(6-2)(6-3)) the answer I got was 180.
This is sufficient eveidence that my theory and formula are all correct.
In conclusion, I think I carried out my investigation accurately. I believe I should have looked into a more consice formula which would have solved all everything. Due to the time constraint I could not do this.
Abdullah Jafar Chowdhury Layers
25/08/2006 Maths