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Introduction

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. ...read more.

Middle

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. ...read more.

Conclusion

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 1 ...read more.

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