14.
a. This problem is simple because it's one-dimensional. It helps you think about how to do the harder problems later on, though.
You have a 3x4 grid, and you need to figure out how many ways there are to fill it with 1x2 blocks. That means there are at most 6 blocks to place. Now we just need to figure out how to place those 6 blocks.
So ask yourself, is there any situation that makes you unable to place a block? Let's start with situations that make blocks really hard to place. If you'll notice, this setup leads to only one possible solution:
![[Image: 3x4.png]](http://img527.imageshack.us/img527/2450/3x4.png)
In fact, as soon as you place that blue block in that spot, you've locked yourself into one solution. Similarly, if you put the red block in the center, you've locked yourself into that one solution. The only other possibility is if the blue block is below the red block, which is the same result, since you can just flip the box 180 degrees. So now let's add one to our total solutions and pretend this one doesn't exist anymore. If we can't place the blue block or the red block, that means you aren't allowed to place any blocks spanning the middle two rows. This is really important. Why? Because now you can cut the box in half down its center, and you won't have to deal with any solutions where a block sticks out halfway from the other side. This makes the problem "how many ways are there to fit 3 1x2 blocks into a 2x3 box?" This is extremely easy to solve. Is there any way you can place a block so that another block can't be placed? No. No matter how you place the first two blocks, the third block will have a place to go. So since there's only three blocks, let's just look for unique situations. Once you place the first block, there are only two ways to place the other two.
![[Image: 3x4.png]](http://img52.imageshack.us/img52/2450/3x4.png)
We can:
1. place all three horizontally (symmetrical) or
2. place one horizontally and have the other two pointing at it (2 facings).
Now we just need to fit the two 2x3 sections together. (1) fits with itself one way, (1) fits with (2) two ways, and (2) fits with itself two ways. That means there are 5 unique solutions.
Add the original solution, and we have a total of 6 for our answer.
I solved this problem extremely methodically, even though it could've been done much faster by guess and check, because it allows you to see how to attempt the other problems, which can't be so easily guess-and-checked. Don't have time to get around to the other parts right now, but hopefully this will help others that do
Edit: Actually, © is ridiculously easy.
You have a 3x3 cube and you have to fill it with 3x1 blocks. These blocks will always fit across the entire cube. What's more, once you place the block that passes through the very center of the cube, you lock yourself into a small set of solutions. We can think of the cube either as a set of three 3x3 squares stacked on top of each other, a three 3x3 squares stacked one behind another. If you think about it, there are no solutions where you can't rotate the cube so that this is true.
I'll choose to start with the cube as 3x3 squares stacked on top of each other. There's only two ways to fit three 3x1 bars into a 3x3 square. Horizontally (H), or vertically (V). Every solution will be based on the facing of the stacked squares. Since the problem states that the top of the box is open, HVV is different than VVH. However, since rotation doesn't matter, HVV is the same as VHH.
There are only four unique solutions.
HHH
HHV
HVH
HVV
Now let's deal with the squares stacked vertically. This time, it's backward. HVV is different than VHH, while HVV is the same as VVH. Also, HHH is an invalid solution, because it's exactly the same as the HHH from the previous part.
Unique solutions:
VVV
VVH
VHH
4 + 3 = 7 solutions.
a. This problem is simple because it's one-dimensional. It helps you think about how to do the harder problems later on, though.
You have a 3x4 grid, and you need to figure out how many ways there are to fill it with 1x2 blocks. That means there are at most 6 blocks to place. Now we just need to figure out how to place those 6 blocks.
So ask yourself, is there any situation that makes you unable to place a block? Let's start with situations that make blocks really hard to place. If you'll notice, this setup leads to only one possible solution:
In fact, as soon as you place that blue block in that spot, you've locked yourself into one solution. Similarly, if you put the red block in the center, you've locked yourself into that one solution. The only other possibility is if the blue block is below the red block, which is the same result, since you can just flip the box 180 degrees. So now let's add one to our total solutions and pretend this one doesn't exist anymore. If we can't place the blue block or the red block, that means you aren't allowed to place any blocks spanning the middle two rows. This is really important. Why? Because now you can cut the box in half down its center, and you won't have to deal with any solutions where a block sticks out halfway from the other side. This makes the problem "how many ways are there to fit 3 1x2 blocks into a 2x3 box?" This is extremely easy to solve. Is there any way you can place a block so that another block can't be placed? No. No matter how you place the first two blocks, the third block will have a place to go. So since there's only three blocks, let's just look for unique situations. Once you place the first block, there are only two ways to place the other two.
We can:
1. place all three horizontally (symmetrical) or
2. place one horizontally and have the other two pointing at it (2 facings).
Now we just need to fit the two 2x3 sections together. (1) fits with itself one way, (1) fits with (2) two ways, and (2) fits with itself two ways. That means there are 5 unique solutions.
Add the original solution, and we have a total of 6 for our answer.
I solved this problem extremely methodically, even though it could've been done much faster by guess and check, because it allows you to see how to attempt the other problems, which can't be so easily guess-and-checked. Don't have time to get around to the other parts right now, but hopefully this will help others that do
Edit: Actually, © is ridiculously easy.
You have a 3x3 cube and you have to fill it with 3x1 blocks. These blocks will always fit across the entire cube. What's more, once you place the block that passes through the very center of the cube, you lock yourself into a small set of solutions. We can think of the cube either as a set of three 3x3 squares stacked on top of each other, a three 3x3 squares stacked one behind another. If you think about it, there are no solutions where you can't rotate the cube so that this is true.
I'll choose to start with the cube as 3x3 squares stacked on top of each other. There's only two ways to fit three 3x1 bars into a 3x3 square. Horizontally (H), or vertically (V). Every solution will be based on the facing of the stacked squares. Since the problem states that the top of the box is open, HVV is different than VVH. However, since rotation doesn't matter, HVV is the same as VHH.
There are only four unique solutions.
HHH
HHV
HVH
HVV
Now let's deal with the squares stacked vertically. This time, it's backward. HVV is different than VHH, while HVV is the same as VVH. Also, HHH is an invalid solution, because it's exactly the same as the HHH from the previous part.
Unique solutions:
VVV
VVH
VHH
4 + 3 = 7 solutions.