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  • Level: GCSE
  • Subject: Maths
  • Word count: 1254

Layers of cubes

Extracts from this document...

Introduction

Part 1

Aim: Investigate how many different arrangements there are for five cubes on the bottom layer when the grid size is 2x3.

Rule 1: The number of cubes on the bottom layer is one less than the number of squares on the grid.

On a grid size of 2x3 squares, there is a possible of 6 different variations using only 5 cubes because all 6 squares have to be empty once. The variations are as follows!

image00.jpg

Part 2

Aim: Investigate the relationship between number of arrangements and the size of the grid when there are:

  1. Two layers of cubes,
  2. More than two layers of cubes

Rule 1: The number of cubes on the bottom layer is one less than the number of squares on the grid.

Rule 2: Each new layer is made with one less cube than the layer underneath it.

image01.jpg

(a)

To find out the formula for 2 layers of cubes, I drew a table, which started from the product of the grid size (G) 3 all the way up to 10. Then in the next column it was the number of cubes on layer 1 (L1), then it was the number of cubes on layer 2 (L2).

...read more.

Middle

6

5

4

210

8

7

6

5

336

9

8

7

6

504

10

9

8

7

720

So the general formula to find out how many arrangements there are for n layers of cubes with 1 less cube on each layer is:

The product of the grid size!

(The product of the grid size – how many layers you want to find out for)!

G!

(G-L)!

L can be equal or less than G

To find out the number of total arrangements, it is the product of the grid size! (G!). L can be equal to G because G is the number of possible variations for the bottom layer.

Part 3

Aim: Investigate the relationship between the number of arrangements and the size of the grid when the number of empty squares on the firstlayer is greater than 1.

Rule 2: Each new layer is made with one less cube than the layer underneath it.

In this part, I am going to investigate the formula for any number of cubes on an X sized grid, with J layers, with Q missing squares on the bottom layer.

image03.jpg

Product of the Grid Size

Number of possible arrangements

3

2

4

6

5

10

6

15

This is a drawing of the bottom layers of 3x1 to 6x1 grid arrangements with 2 squares missing.

This is a drawing of the bottom layers of 4x1 to 7x1 grid arrangements with 3 squares missing.image04.jpg

Product of the Grid Size

Number of possible arrangements

4

4

5

10

6

20

7

35

This is a drawing of the bottom layers of 5x1 and 6x1 grid arrangements with 4 squares missing. image05.jpg

Product of the Grid Size

Number of possible arrangements

5

5

6

15

7

35

8

70

...read more.

Conclusion

8!

   (8-4)! 4!

For the layers on top of that I drew up the following 2 tables:

2 Squares Missing

Product

Bottom Layer

Layer 2

Layer 3

Layer 4

3

2

4

6

2

5

10

3

2

6

15

4

3

2

7

21

5

4

3

3 Squares Missing

Product

Bottom Layer

Layer 2

Layer 3

Layer 4

4

4

5

10

2

1

6

20

3

2

1

7

35

4

3

2

8

56

5

4

3

When there are 2 squares missing if I want to go up to the second layer, it is G-2, to go up to the third layer it is G-2-1. When it 3 squares for layer 2 it is G-3 and for layer 3, it is G-3-1.

After I drew this table, I knew it had something to do with the general formula for the bottom layer,

G!

   (G-M)! M!

So that had to be multiplied by something because I wanted to find how many variations I could have when G=6, L=2 and M=2. The answer had to be 60.

6!

   (6-2)! 2!

I also knew that in the next part of the formula, the number of layers you wanted to find out for had to be included, as well as the number of missing squares.

Through trial and improvement I got the next part to be

      (G-M)!

   (G-L-M+1)!

To test my formula I used G=6, L=2 and M=2, and it worked.

G!                                

                    (G-M)! M!                (G-L-M+1)!

6!                                

                    (6-2)! 2!                        (6-2-2+1)!

Then I realized that The (G-M)! Could be canceled out, to leave the formula at:

G!                                

                            M!                        (G-L-M+1)!

...read more.

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