(b)
To find out the formula for more than 2 layers, I drew the same table again… the first part is the same, 5 cubes can be arranges on a grid size of 6 squares 6 times, and then 4 cubes can be arranges on 5 cubes 5 times, and then 3 cubes can be arranged on 4 cubes 4 times and so on… so what I did was I multiplied 6 by 5 to get 30 (which was how many arrangements there are for 2 layers) , then I multiplied by 4 to find out the arrangements for 3 layers and it came up to 120 arrangements.
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 first layer 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.
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.
This is a drawing of the bottom layers of 5x1 and 6x1 grid arrangements with 4 squares missing.
I didn’t complete this diagram because I was beginning to see a pattern. With 2 squares missing, the formula Gx(G-1)÷2. When there was 3 squares missing, the formula was Gx(G-1)x(G-2)÷6, and when there was 4 missing squares, the formula was Gx(G-1)x(G-2)x(G-3)÷24.
It was then I realized the pattern. I knew the formula would have something to do with
The Product of the Grid size!
G!
To prove my formula I used G=5 and M=2. So 5!÷2!=60, and from the tables I knew the answer had to be 10. The only way to get 60 down to 10 was dividing by 6. Since the formula is factorial, I used 3! Then I tried it again… and it worked. The only way to get 3 from 5 is to minus 2, which is M. so the general formula would be
G!
(G-M)! M!
To prove my formula I took G to be 8 and M to be 4.
8!
(8-4)! 4!
For the layers on top of that I drew up the following 2 tables:
2 Squares Missing
3 Squares Missing
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)!
