LEGO R

bricks are very common toys and have a structure that allows them to be placed
on a rectangular grid. This makes them great for interesting questions about math. A 2�4
brick has two rows and four columns, making it have 8 total studs. But a 2�4 is the same
as a 4 � 2, just rotated 90 degrees.
In this assignment, you are going to �nd how many ways you can put a 2 � 4 on top
of another 2 � 4. We will give some simple examples of what this means. If we have a
1�1, there are 8 places on the top of a 2�4 we can place it if we don't move or rotate the
bottom. But, some of these positions are going to be the same. If I place a 1�1 in the top
left corner of a 2 � 4, and rotate the 2 � 4 by 180�, we see that it is the same as placing a
1 � 1 in the bottom right of the 2 � 4.
So, there are only four ways to put a 1 � 1 on a 2 � 4, since placing a 1 � 1 in any spot
on the top row will be moved to a spot on the bottom row by rotating the whole brick. If
we used a 1 � 2 instead of a 1 � 1, we would have more ways, since the brick can hang o�
the side. Picture below are the only two ways to place a 1 � 2 on the top left stud.
For this assignment, you will:
(1) Find the number of positions you can place a 1�2 on top of a 2�4 without moving
or rotating the bottom 2 � 4. (You can rotate the 1 � 2, but only by 90�.) Be sure
justify why you found every way and there are no others.
(2) Use part (1) to �nd out how many ways you can place a 1 � 2 on top of a 2 � 4 if
you are now allowed to rotate the bottom brick after you place the top brick. Be
sure justify why you found every way and there are no others.
(3) Instead of using a 1 � 2 to put on top of a 2 � 4, use a 2 � 4 to:
(a) Repeat part (1). Be sure to justify your solution as you did in part (1).
(b) Repeat part (2). Be sure to justify your solution as you did in part (2).

(1) Finding the number of positions you can place a 1x2 brick on top of a 2x4 brick without moving or rotating the bottom 2x4:

To solve this, we need to consider the possible positions of the 1x2 brick on top of the 2x4 brick. Since the 1x2 brick can only be rotated by 90 degrees, there are limited possibilities.

Let's assume the 2x4 brick is represented by a rectangular grid with 2 rows and 4 columns:

```
_ _ _ _
_ _ _ _
```

We can place the 1x2 brick in two orientations:

Orientation 1 (vertical orientation):
```
X _ _ _
X _ _ _
```

Here, the "X" represents the 1x2 brick placed vertically. There are three possible positions where this orientation can be placed on top of the 2x4 brick.

Orientation 2 (horizontal orientation):
```
X X _ _
_ _ _ _
```

Here, the "X" represents the 1x2 brick placed horizontally. There are two possible positions where this orientation can be placed on top of the 2x4 brick.

Therefore, the total number of positions to place the 1x2 brick on top of the 2x4 brick without moving or rotating the bottom brick is 3 (vertical orientation) + 2 (horizontal orientation) = 5.

(2) Finding the number of positions you can place a 1x2 brick on top of a 2x4 brick if you are allowed to rotate the bottom brick after you place the top brick:

In this case, we are allowed to rotate the bottom brick after placing the top brick. This means that there will be additional possibilities for placing the 1x2 brick.

We can still consider the two orientations mentioned in part (1):

Orientation 1 (vertical orientation):
```
X _ _ _
X _ _ _
```

Similar to part (1), there are three possible positions for this orientation.

Orientation 2 (horizontal orientation):
```
X X _ _
_ _ _ _
```

However, in this case, we can also rotate the bottom brick by 180 degrees, which allows us to have two more positions for this orientation.

Therefore, the total number of positions to place the 1x2 brick on top of the 2x4 brick, considering the ability to rotate the bottom brick, is 3 (vertical orientation) + 4 (horizontal orientation) = 7.

(3a) Using a 2x4 brick to place on top of a 2x4 brick:

We can apply the same logic as before. Since the 2x4 brick cannot be rotated, there is only one possible way to place it on top of the 2x4 brick - aligning it perfectly.

Therefore, there is only one way to place a 2x4 brick on top of a 2x4 brick without moving or rotating the bottom brick.

(3b) Using a 2x4 brick to place on top of a 2x4 brick, considering the ability to rotate the bottom brick:

In this case, we can rotate the bottom brick after placing the top brick, which allows for additional possibilities.

Since the 2x4 brick is the same size as the bottom brick, rotating it would not change its position. Therefore, there is still only one possible way to place it on top of the 2x4 brick, regardless of whether rotation is allowed or not.

Therefore, there is also only one way to place a 2x4 brick on top of a 2x4 brick, considering the ability to rotate the bottom brick.

To find the number of positions you can place a 1x2 brick on top of a 2x4 without moving or rotating the bottom 2x4, you need to carefully analyze the placement options. Let's break down the problem step by step.

Step 1:
Consider the 2x4 bottom brick and label each stud (small circular part) with numbers from 1 to 8, starting from the top left and moving left to right, then top to bottom. This will help us visualize and refer to each position more easily.

Step 2:
Now, let's place the 1x2 brick on top of the 2x4 brick. Since the 1x2 brick can only be rotated by 90 degrees, we have two possible orientations for it: vertical and horizontal.

Step 3:
For the vertical orientation, place the 1x2 brick on the leftmost stud of the bottom brick, covering two consecutive studs. Let's analyze this placement:
- If we place the 1x2 brick on the first stud, it covers studs 1 and 2.
- Similarly, if we place it on the second stud, it covers studs 2 and 3.
- If we move it to the third stud, it covers studs 3 and 4.
- Finally, if we place it on the fourth stud, it covers studs 4 and 5.

So, for the vertical orientation, there are four possible positions to place the 1x2 brick. These are determined by the leftmost stud's position.

Step 4:
For the horizontal orientation, place the 1x2 brick on the topmost studs of the bottom brick, covering two consecutive studs. Again, let's analyze this placement:
- If we place the 1x2 brick on the first stud, it covers studs 1 and 6.
- Similarly, if we place it on the second stud, it covers studs 2 and 7.
- If we move it to the third stud, it covers studs 3 and 8.

For the horizontal orientation, there are three possible positions to place the 1x2 brick. These are determined by the topmost studs' position.

Step 5:
Combining the results from the vertical and horizontal orientations, we find a total of seven possible positions to place the 1x2 brick on top of the 2x4 without moving or rotating the bottom brick.

Moving on to the second part of the problem:

Step 6:
Now, we need to find the number of ways to place the 1x2 brick on top of a 2x4 if we are allowed to rotate the bottom brick after placing the top brick.

Step 7:
For the vertical orientation, we have already considered all possible positions in the previous part. No additional positions are possible since rotating the bottom brick wouldn't change the covered studs.

Step 8:
For the horizontal orientation, rotating the bottom brick allows for an additional possibility. If we place the 1x2 brick on the first stud and then rotate the bottom brick by 180 degrees, the covered studs would be 4 and 3, which is a new position not accounted for before.

Therefore, for the second part, there are a total of eight possible positions to place the 1x2 brick on top of the 2x4 when allowing rotation of the bottom brick.

Now let's move on to using a 2x4 brick instead:

Step 9:
For the first part, using a 2x4 brick on top of a 2x4, we need to find the number of positions without moving or rotating the bottom brick.

Step 10:
Since the 2x4 brick already covers the entire top surface of the bottom brick, there is only one possible position, which is placing the 2x4 brick directly on top of the 2x4 bottom brick.

Step 11:
For the second part, where we can rotate the bottom brick after placing the top brick, the number of possibilities remains the same. Even with rotation, there is only one feasible position. Since the 2x4 brick already fills the entire top surface, rotating it wouldn't change the covered area.

To summarize:

(1) For a 1x2 on top of a 2x4 without moving or rotating the bottom, there are 7 positions.
(2) For a 1x2 on top of a 2x4 with the ability to rotate the bottom, there are 8 positions.
(3) For a 2x4 on top of a 2x4 without moving or rotating the bottom, there is only 1 position.
(4) For a 2x4 on top of a 2x4 with the ability to rotate the bottom, there is still only 1 position.

Remember to justify your solutions by explaining the thought process and reasoning behind the placements.