Mathematics Layers Coursework

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Mathematics Layers Coursework

Chris Grindrod  11SRH

Introduction:

In this investigation I will be investigating the amount combinations there are when there is a set sized grid with x number of squares on it, one cube can fit In to each square but there must also be one empty square. Then another layer of squares can be placed on top of the first layer of squares but the same rules apply and there must be one square that does not have another square placed on top of it. This applies to any number of layers until you are made to or required to stop. On each new layer you have to use one cube than the layer before until you have to stop or are asked to stop at a certain layer.

The rules of the investigation are as follows:

The number of cubes on the bottom layer is always one less than the number of squares on the grid unless stated otherwise.

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

Each cube must fit exactly in to a square or a cube underneath it.

Part One

For part one of the investigation I will investigate how many different arrangements of five cubes there are on a six square grid.

Method:

I am going to draw all the combinations for the specifications above to find out how many combinations there are.

Results:

There are six combinations for the specifications, this is because no matter were you put the five cubes on the six square grid there will always be one vacant square and because there are six squares on the grid you can have six combinations, one for each square.

Algebra:

The formula for this problem is:

n

This is were n equals the size of the grid. When we substitute numbers in it is:

 6

Part Two A

For part two A I will investigate the relationship between number of arrangements and the size of the grid when there are two layers of cubes on a grid of six squares.

Method:

I am, like part 1, going to draw out all of the combinations for the specifications above to find out how many combinations there are.

Results:

(next page)

There are thirty combinations because, like the last one, there is always a space left over on the bottom layer so there will be six combinations for that, and on the next layer it is the same principal but with a an entire space missing were there are no spaces or cubes. So on the next layer there is going to be a space missing with each combination for five combinations, so it will just be six multiplied by five because the first layer has six combinations and the second layer has five combinations.

Algebra:

In this formula we need to write six multiplied by five, instead of writing this like 6´5 I am going to write it in a different way that will be useful later on in this investigation. I am going to use factorial numbers, which means putting an exclamation mark next to the number,

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E.g.

6!

This basically means that this number, and every number up to it, is multiplied together.

E.g.

6! = 6x5x4x3x2x1

If I want to write 6´5 using factorial numbers you cant write six factorial because then you are multiplying it by four and three and two and one as well and you don’t need those extra numbers. To get over this problem you have to divide six factorial by four factorial so that you can cancel out the unwanted numbers.

E.g.

Now I have to incorporate this in to a ...

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