• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Mathematics Layers Coursework

Extracts from this document...


Mathematics Layers Coursework

Chris Grindrod  11SRH


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.


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



...read more.


Now I will choose a number of layers to do, I have chosen to use a grid size of four with two layers because I have already have the number of combinations (12), which is my target answer, and in the table above I have the answers with which to check my answer.  image22.pngimage23.png

I know that I have to devide 4!  by something because we are treating each number in 4! as a new layer, and because the 4! Is greater than 12 I know that there are too many layers involved. The only direct way to get rid of any numerator is to devide by the same number, so if I want the first two layers (4x3) I must devide by the remaining layers (2x1).image24.png


Now I shall do the same with three layers, the number of combinations for three layers on a four grid is 24, 24 is my target answer.


Now I shall do the same with one layer, the number of combinations for one layer on a four grid is 4, 4 is my target answer.

I will now put this information in to a grid.image03.png

Grid Size (n)

Layers (r)





3x2x1 or (3!)




2x1 or (2!)




1 or (1!)


My table shows me that, if you think of each number in the denominator column as a layer;

I.e. 3x2x1 (or 3!) is three layers because there are three numbers there.

Then the number of layers plus the number of

...read more.


n) is mentioned in this part of the formula, it must be subtracted by the number of gaps on the bottom layer (o). So the formula now reads:


I have to multiply these two parts of the formula together because one half caters for what effect the number of gaps has on the bottom layer and the half above caters for the problem of any sized grid and any number of layers.

. So the formula for part three will be:




It is this answer because:


My formula for the combinations of a four square grid with two layers of cubes and two gaps on the bottom row gives me the same answer as my results, which means that my formula for part three is correct.

...read more.

This student written piece of work is one of many that can be found in our GCSE Number Stairs, Grids and Sequences section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Number Stairs, Grids and Sequences essays

  1. Number Grid Coursework

    A graph is a good choice to show this relationship however, a straight line drawn between these would not be correct because with this particular problem, the values to be inputted into equations must be natural numbers i.e. Integers > 0.

  2. Number Grids Investigation Coursework

    (n - 1) would equate to (n - 1)2 in a square. I will check this formula by trying a 3 x 4 rectangle, which, according to my formula would have a difference between the products of opposite corners of: D = 10 (m - 1)

  1. Algebra Investigation - Grid Square and Cube Relationships

    both answers begin with the equation n2+22n, which signifies that they can be manipulated easily. Because the second answer has +40 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 40 will always be present.

  2. Number Grid Investigation.

    - (24 X 72) = 100. The product difference increases by 20 as the depth of the square increases by 1. Size of square 3 X 4 3 X 5 3 X 6 ... 3 X 7 3 X 8 Product difference 60 80 100 ...

  1. Maths - number grid

    36 9 x4 9 x 2 4 x4 81 9 x 9 9 x 3 5 x5 144 9 x 16 9 x 4 I can now draw upon the conclusion that the formula for this sequence of rhombuses is (10-1)(r-1)

  2. Maths-Number Grid

    56 57 37 38 39 40 47 48 49 50 57 58 59 60 67 68 69 70 I have learnt from the above 4 � 4 grids that the difference between the products has not changed throughout the following 3 cases I have completed.

  1. Number Grid Maths Coursework.

    26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75

  2. Maths coursework. For my extension piece I decided to investigate stairs that ascend along ...

    This shows that my formula must work for all 4-stair shapes on the 5 x 5 grid. To investigate if this formula also works for 4-stair shapes on other grid sizes, I am going to see if the formula still works for 4-stair shapes on a 7x 7 grid.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work