I decided to finish up 14 while I'm at it.
14.
b.
![[Image: 2x2x3.png]](http://img52.imageshack.us/img52/7670/2x2x3.png)
Once again, there are 6 blocks to place. The shaded 2x2 area must be filled with two complete blocks. If you were to lay any block along the x-axis instead, you could just rotate the box 180 degrees, and the other side would have to be filled with 2 blocks. There are 2 ways to fill a 2x2 area with 2 blocks.
Now let's look at the rest of the box. We have a 2x2x2 cube, and 4 blocks to fill it with. This is simple enough to start listing possiblilities.
1. We can stack them all vertically.
2. We can stack them all horizontally, facing the x-axis.
3. We can have the ones in front stand vertically, and the ones in the back stand horizontally.
4. We can have the ones in front stand horizontally, and the ones in the back stand vertically.
That's 8 possibilities.
We have to be careful with setups where no block lies along the x-axis, because some of the possibilities are duplicated when you flip the box around. The middle row of the box is unaffected by rotation, so there are two possibilities there. There are three ways to set up the left and right ends of the box.
1. Left vertical, right horizontal
2. Both vertical
3. Both horizontal
That's 6 possibilities.
Answer is 8 + 6 = 14 solutions.
Will edit when I solve the rest.
d.
This problem is similar to part (a). From part (a), you can see that if you stand a block up so that it takes up a 1x2 area on the 3x4 grid, there are 6 possibilities. Sweet, halfway done. All our other solutions must contain at least one block that lies on its side.
I'm going to proceed in the same way as I did in part (a). Figure out which solutions involve the middle of the box, eliminate them, and cut the box in half. You can see that you are not allowed to completely fill up the middle of the bin with 3 blocks, because you can't divide 3x3 areas into 3x2 sections. Say we place a block floating in the center of the bin. There are now only two ways to fill the rest of the bin.
![[Image: 3x3x4.png]](http://img693.imageshack.us/img693/7893/3x3x4.png)
The other is the same thing, upside down. There are no other ways to fill the bin with blocks spanning the center. So now that we have our two solutions, we can cut the box in half.
Ways to fill a 3x3x2 box with 3x2x1 blocks not counting standing all blocks up:
1. 3 stacked on top of each other.
2. One flat on the bottom, other two vertical
3. Two vertical, one flat on the top.
3 choose 2 = 6 permutations.
Answer is 6 + 2 + 6 = 14 solutions.
e. There are 9 blocks in this one. Luckily, it is impossible to fit all the blocks into the given area. The answer is 0.
f. Not too hard to see that once you place the yellow block, there is only one solution.
![[Image: 2x2x32.png]](http://img714.imageshack.us/img714/298/2x2x32.png)
All other solutions involve the blocks all lying flat or all standing up and facing the same direction. There are 4 other solutions, for a total of 5.
g. I only found 2 solutions via trial and error. I'm pretty sure there aren't any more, though.
For convenience, solutions to all parts of problem 14 here:
a. 6
b. 14
c. 7
d. 14
e. 0
f. 5
g. 2
14.
b.
Once again, there are 6 blocks to place. The shaded 2x2 area must be filled with two complete blocks. If you were to lay any block along the x-axis instead, you could just rotate the box 180 degrees, and the other side would have to be filled with 2 blocks. There are 2 ways to fill a 2x2 area with 2 blocks.
Now let's look at the rest of the box. We have a 2x2x2 cube, and 4 blocks to fill it with. This is simple enough to start listing possiblilities.
1. We can stack them all vertically.
2. We can stack them all horizontally, facing the x-axis.
3. We can have the ones in front stand vertically, and the ones in the back stand horizontally.
4. We can have the ones in front stand horizontally, and the ones in the back stand vertically.
That's 8 possibilities.
We have to be careful with setups where no block lies along the x-axis, because some of the possibilities are duplicated when you flip the box around. The middle row of the box is unaffected by rotation, so there are two possibilities there. There are three ways to set up the left and right ends of the box.
1. Left vertical, right horizontal
2. Both vertical
3. Both horizontal
That's 6 possibilities.
Answer is 8 + 6 = 14 solutions.
Will edit when I solve the rest.
d.
This problem is similar to part (a). From part (a), you can see that if you stand a block up so that it takes up a 1x2 area on the 3x4 grid, there are 6 possibilities. Sweet, halfway done. All our other solutions must contain at least one block that lies on its side.
I'm going to proceed in the same way as I did in part (a). Figure out which solutions involve the middle of the box, eliminate them, and cut the box in half. You can see that you are not allowed to completely fill up the middle of the bin with 3 blocks, because you can't divide 3x3 areas into 3x2 sections. Say we place a block floating in the center of the bin. There are now only two ways to fill the rest of the bin.
The other is the same thing, upside down. There are no other ways to fill the bin with blocks spanning the center. So now that we have our two solutions, we can cut the box in half.
Ways to fill a 3x3x2 box with 3x2x1 blocks not counting standing all blocks up:
1. 3 stacked on top of each other.
2. One flat on the bottom, other two vertical
3. Two vertical, one flat on the top.
3 choose 2 = 6 permutations.
Answer is 6 + 2 + 6 = 14 solutions.
e. There are 9 blocks in this one. Luckily, it is impossible to fit all the blocks into the given area. The answer is 0.
f. Not too hard to see that once you place the yellow block, there is only one solution.
All other solutions involve the blocks all lying flat or all standing up and facing the same direction. There are 4 other solutions, for a total of 5.
g. I only found 2 solutions via trial and error. I'm pretty sure there aren't any more, though.
For convenience, solutions to all parts of problem 14 here:
a. 6
b. 14
c. 7
d. 14
e. 0
f. 5
g. 2