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 formula. I know that the numerator is the grid size and I got to using this denominator because there was four numbers I didn’t need, but I had to leave two numbers that I did need because those numbers were the combinations for the first two layers. So to get the denominator you have to subtract the number of layers from the grid size then use it as a factorial number. So the formula is:
This formula works were n equals the grid size and r equals the amount of layers. When we substitute in numbers it is:
Part Two B
For part two B I will investigate the relationship between number of arrangements and the size of the grid when there are more than two layers of cubes
Method:
I will look at previous formulas to try and see if they have any relevance to this problem and I will adapt them. I will then develop a formula through mathematical logic and test it against some results that I will draw.
Results:
(next page)
In my previous formula for Part Two A I used a formula that involved both grid size and number of layers to get the right number of combinations. The formula was;
In this formula I have the grid size as a factorial, which will give us the total number of combinations with all the layers involved because factorial means all the numbers up to and including the one I have chosen multiplied together.
That will give us the right number of combinations for the problem if there is one gap on the bottom row and we need the number of combinations for all layers. Our problem, however, is that I must find a formula with any number of layers which means that the number of layers, or r, must be incorporated in to my formula because that is the variable that will change the number of combinations.
Unless we use all layers, n! will be too large a number because that will mean the number of combinations for all the layers. We can find out what we must do to manipulate n! by doing the following;
Replace n with a number, in this instance four because I have it is large enough to give enough varied results with different layers but not too big as to take up too much time drawing the combinations.
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.
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).
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.
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 layers the denominator represents equals the grid size, which means that the number of layers, or r, must feature in the final formula because r effects the size of the denominator, or how much we have to divide the numerator by. Because the denominator is what is important in the formula I need to write part of a formula to represent it using the grid size and the number of layers.
What I am trying to achieve with this part of the formula is to manipulate the numbers so that I can get a formula that gets the correct number of layers. Because the grid size, the number of layers and the number of layers the denominator represents are all related (Number of layers + Amount of layers in the denominators = grid size) I can safely say that the grid size (n) subtract the number of layers (r) equals the number of numbers in the denominator or the number of layers in the denominator or, in other words, how many layers I do not need. All the numbers must be made factorial because we are thinking of consecutive layers counting down. So;
This will equal the grid size as a factorial by the number of layers I do not need as a factorial.
I know that the denominator will be a factorial because any numbers in it will be in consecutive order and will be multiplied together as shown in my table above.
Now I know how to make the numerator and the denominator I can produce a formula.
Now I have to check it by replacing the letters with numbers.
That is the answer because;
This formula works for a four square grid with three layers.
This formula works for a four square grid with two layers.
I have come to the conclusion that my formula works.
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.
Method:
I am going to use grids that are relatively small grid so that it is possible to draw a;; the combinations, then I am going to formulate a formula that works and test it.
Results:
(next page)
Algebra:
To explain this properly I will need an example from my results:
A four square grid,
Two layers,
Two blocks missing from the bottom layer.
I will use n as the grid size, o as the amount of gaps on the bottom layer and r as the amount of layers.
For my example I am going to use grid size four, with two gaps on the bottom row and two layers. This gives me twelve combinations.
For the number of places where the gaps can go is four spaces on the grid for the first gap, then three spaces on the grid for the second gap, so for the number of places the gaps could go could be written as:
4x3
This is nearly four factorial but four factorial would be this:
4x3x2x1
But we don’t need to multiply it by two and one so we need to divide it by two and one (or two factorial) so we can cancel out the unwanted numbers.
The amount we divide the numerator by is governed by how many layers there are because the more layers we need, the less numbers we need to cancel out. So we divide the numerator (the grid size as a factorial) by the grid size subtract the amount of layers. We do this because, supposing we use all the layers available to use after we have taken away x number of gaps on the bottom row, the number of layers plus the number of gaps equals the grid size.
This can be proved by saying the we have a grid size of six, there will be five layers because the first layer will have five blocks and the following layers will get consecutively smaller by one block. If we have five blocks on the bottom layer on a size six grid, then there will be one gap on the bottom row, so:
Going back to our example. Now that I have explained why, to make the denominator, we must subtract the number of layers away from the grid size now we have to we make that a factorial as well, because we wont just need to cancel out one number, we will need to cancel out every number up until that number. Here is the example:
Or
But the problem with this is that we have effectively designated each gap on the grid in the order that we chose them and we don’t need to specify the order, so we have over-counted. We have over counted because, say we have two spaces for gaps, we have counted in a way that puts one gap in the first space, then the second gap in the second space, but then we have reversed these so the first is in the second and the second is in the first.
These two ways of counting achieve the same result, which means we have over counted and will need to reduce the number of combinations for the places were the gaps go. To over come this problem we must multiply the denominator by the amount of gaps as a factorial because, in this case, two factorial is the number of ways of counting the number of gaps, you can give any one of the two gaps the first space and any one the second, so we have to multiply it by two and one or two factorial. So the formula, with the numbers still substituted in, now reads:
Or
This is not the whole formula because this only caters for the problem with any number of gaps on the bottom layer so we must try and find the next part of the formula that will include the number of layers and the remaining grid size after the gaps have been included. This part of the formula will not be dissimilar from the previous formula of:
In the formula above I was trying to find out the amount of combinations with any number of layers and any sized grid, that is what I am trying to find for part three except we need to include the fact that we have to be able to take away any number of gaps on the bottom layer. I have already made part of the final formula that will account for the main alterations that having any number of gaps on the bottom layer will make. Now I have to alter the formula above to account for any number of gaps on the bottom layer because this half of the formula accounts for the number of layers and the grid size so the number of gaps on the bottom layer is very important to this part of the formula. To overcome this problem I am going to use the formula above, but with the number of gaps taken away from wherever the grid size features. I am only subtracting it from the grid size and no the number of layers because, although it does play a huge part in the number of layers, the grid size features wherever the number of layers features, so I have no need to subtract it from both. So wherever the grid size (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:
Or
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.