It is a grade 12 course, and our current unit is Combitionations. There is on questions where there is a 4 x 4 cubic diagram, like the 1x1 squares on graphing paper but 4x4.
The question is to explain how the formula 5C2 x 5C2 gives the answer to how many rectangles (or simply quadrilaterals) can be formed in this box. The formula leads to the answer 100, but I don't know how to explain how it works in relation to the diagram.Thanks
You'll need to derive the formula yourself.
Start with 1x1.
Then deduce a method for finding 2x2 and 3x3.
Generalize your formula.
You may want to use induction.
Good luck.
Look at the grid across the top and down the side. If you start in the upper left hand corner (it doesn't matter where you start, but this is natural) then you'll see that there are 5 vertices across the top. A similar observation hold if you go down the far left side. Thus there are 5C2 ways to pick the 2 top vertices and a similar result holds for the vertices on the left side. The conclusion should follow once you see how the vertices are being chosen.
Try to formulate a general result for a rectangular grid that contains m x n square cells in it.
After reviewing my post I think it would be a little clearer to forget the selection as being for vertices. It might help to think of selecting lines to form the perimeter of each rectangle. There are 5 vertical lines and 5 horizontal lines that make up the 4x4 grid. We need to select 2 of each, so there are 5C2 ways to select vertical lines and the same number of horizontal lines.
Hopefully this is clearer than my first post.
The two expressions that give you your answer are found in the following.'
Lets look at squares first to show you the process.
If you determine the number of squares on smaller boards starting with one square you will readily discover a pattern that leads to a simple formula for a board of any number of squares.
A one square board obviously has only one square.
A 2x2 square board has 5 squares, the 4 basic ones and the one large 2x2 one.
A 3x3 square board has 14 squares, the smaller 9 plus 4 2x2's plus 1 3x3 one.
A 4x4 square board has 30 squares, the smaller 16 plus 9 3x3's, plus 4 2x2's plus 1 4x4 one.
Are you beginning to see the pattern?
1x1 - 1^2 = 1
2x2 - 2^2 + 1^2 = 5.
3x3 - 3^2 + 2^2 + 1^2 = 14.
4x4 - 4^2 + 3^2 + 2^2 + 1^2 = 30.
5x5 - 5^2 + 4^2 + 3^2 + 2^2 + 1^1 = 55.
What would your guess be for the number of squares on an 8 x 8 board? Can you derive a general expression for the answer?
Here it is. Let N(s)n = the total number of squares in a square of nxn squares. Then
N(s)n = n^2 + (n-1)^2 + (n-2)^2................(n-n+1)^2 = n(n + 1)(2n + 1)/6
.
So for the typical chess board problem with 8x8 squares, the total number of definable squares is
N(s)8 = 8(8 + 1)(16 + 1)/6 = 204
Now for how many rectangles there are in a square nxn squares big? We count only "rectangles", not the squares which are special cases of rectangles. Remember, only rectangles where the length is longer than the width.
Again, the best way to creep up on a solution is to start out with the small size squares and see where it leads you. For a 2x2 square, we have a total of 4 possible rectangles, each 1x2. For the 3x3 square, we can find 12 1x2 rectangles, 6 1x3 rectangles, and 4 2x3 rectangles for a total of 22. For the 4x4 square, we can find 24 1x2's, 16 1x3's, 8 1x4's, 12 2x3's, 6 2x4's, and 4 3x4's for a total of 70 rectangles. For the 5x5 square, we get 40 1x2's, 30 1x3's, 20 1x4's, 10 1x5's, 24 2x3's, 16 2x4's, 8 2x5's, 12 3x4's, 6 3x5's, and 4 4x5's.
Did you notice anything as you looked through these numbers? Lets put them into a tabular form. Along side each nxn square will be the number of rectangles identified at the top of the column.
1x2 1x3 1x4 1x5 2x3 2x4 2x5 3x4 3x5 4x5 Total
2x2 4 4
3x3 12 6 4 22
4x4 24 16 8 12 6 4 70
5x5 40 30 20 10 24 16 8 12 6 4 170
NOTICE ANYTHING???
1--The number of 1 square wide rectangles in each case is equal to n^2(n-1).
2--The number of 2 square wide rectangles in each case is equal to the number of 1 square wide rectangles in the previous case. The number of 3 square wide rectangles in each case is equal to the number of 2 square wide rectangles in the previous case.
The total number of rectangles in a square of nxn squares is equal to the sum of the 1 square wide rectangles for each rectangle from the 2x2 up to and including the nxn one being considered. Thus, the number of rectangles in a 5x5 square is the sum of the 1 square wide rectangles in the 1x1, 2x2, 3x3, 4x4, and 5x5 squares or 4 + 18 + 48 + 100 = 170. Therefore, for the typical chess board problem of 8x8 squares, we have a total of 4 + 18 + 48 + 100 + 180 + 294 + 448 = 1092.
Letting N(r)n = the total number of rectangles in a square of nxn squares, we have
N(r)8 = n(3n^3 + 2n^2 - 3n - 2)/12 = 8[3(512) + 2(64) - 3(8) - 8]/12 = 1092.
If you want to derive the total number of squares and rectangles, then the total on an 8x8 checkerboard would be 204 + 1092 for a total of 1296, the square of 36.
Letting N(s,r)n = the total number of squares and rectangles in a square of nxn squares, we have
N(s,r)n = n(n + 1)(2n + 1)/6 + n(3n^3 + 2n^2 - 3n - 2)/12 = [(n^2 + n)^2]/4
For our 8x8 board, N(s,r)8 = [(n^2 + n)^2]/4 = [(64 + 8)^2]/4 = 1296
This is making things unnecessarily conplex. There are 5 choose 2 ways to select vertices for the top and 5 choose 2 ways to choose them for the side. That product gives the total number of rectangles that can be chosen from the grid.
If we had an m x n rectangular grid of squares, then there would be m+1 choose 2 ways to select the top vertices and n+1 choose 2 ways to select the side ones.
I need help solving combination equations.
a tree farm advertises that 25% of their saplings are taller than 1.5m and 10% of them are taller than 2m.
a- what is the mean height of the tree saplings?
b- what is the standard deviation of the heights?
c- in a sample of 60 trees, how manyy would be taller than 2.2m?
I have just started grade 12 U Data Management, and i do not understand tree diagrams. I need help with the following question: Draw a tree diagram representing the playoffs of eight players at the Australian open tennis tournament.
To solve this problem, you can use the concept of probabilities and tree diagrams.
a) Mean height of the tree saplings:
To find the mean height, we need to calculate the weighted average of the heights. Given that 25% of the saplings are taller than 1.5m and 10% are taller than 2m, we can assign weights of 0.25 and 0.10 to their respective heights. The mean height is then calculated as follows:
Mean height = (0.25 * 1.5m) + (0.10 * 2m)
= 0.375m + 0.2m
= 0.575m
b) Standard deviation of the heights:
To calculate the standard deviation, we need additional information such as the variance or the range of the heights. If you have that information, you can use the appropriate formulas to calculate the standard deviation. If not, you might need to clarify the question or seek additional information.
c) Number of trees taller than 2.2m in a sample of 60 trees:
To determine the number of trees taller than 2.2m in a sample of 60 trees, we need to know the overall proportion of trees that are taller than 2.2m in the tree farm. Without this information, it is not possible to provide a specific answer.
Regarding your question about drawing a tree diagram representing the playoffs of eight players at the Australian Open tennis tournament, the concept of tree diagrams is used to represent possible outcomes and decision points in a sequential manner. However, without specific details or criteria regarding the playoffs, it is not possible to provide a specific tree diagram for this scenario. Can you please provide more details or criteria for the playoffs?