B Arranging blocks so that they fit together

[thebosonbreaker]

I have attached an image showing a (what I believe to be) simple problem involving arranging four blocks, each of different dimensions.
Yes, the blocks fit together perfectly in the first arrangement shown in the diagram when there are no gaps.

I'm convinced that the solution is likely easily arrived at, but I am neither confident nor experienced in any in permutations and combinatorics in general.
I believe that the answer is likely to come from the use of some sort of formula involving factorials and all that good stuff but I'm not entirely sure how.

Can anyone give me an idea of where to begin?
Or am I totally incorrect in approaching this problem as one involving combinatorics?
Thank you very much for your help in advance.

 

[scottdave]

Here is one thought. First, we are not given any measurements, so we don't know if the blue and green could be moved above the orange to fill in that same width.
Let's say that is not possible. So they need to stay in the same orientation. I will call the thin horizontal block, which looks Peach colored, P, then the one above it O (for orange), then B for blue and G for green.

So starting with P, you have 2 choices, O can be above it or below it. Then that 2-block group can have the on the left of it, or the right. Then next you can put [G] to the left or to the right of this group. But then there is the case of putting the green first and then choosing blue 2nd. There will be some overlap as rotation could produce duplicates. These are just some ideas to get you thinking about it, though.

 

[thebosonbreaker]

Here is one thought. First, we are not given any measurements, so we don't know if the blue and green could be moved above the orange to fill in that same width.
Let's say that is not possible. So they need to stay in the same orientation. I will call the thin horizontal block, which looks Peach colored, P, then the one above it O (for orange), then B for blue and G for green.

So starting with P, you have 2 choices, O can be above it or below it. Then that 2-block group can have the on the left of it, or the right. Then next you can put [G] to the left or to the right of this group. But then there is the case of putting the green first and then choosing blue 2nd. There will be some overlap as rotation could produce duplicates. These are just some ideas to get you thinking about it, though.

Yes, I have been trying to figure out the solution using this approach. I assume that B and G cannot fill in the width above O. The trouble is that I am in constant fear of missing out one (or more) of the possibilities and therefore arriving at an incorrect answer.
I am more interested in whether or not there is some sort of definitive method of deducing the number of possible arrangements using a more formulaic method.
Nevertheless, it seems simple enough to solve using the visual approach as long as everything is triple-checked.
Thanks for the input :)

 

[Stephen Tashi]

I am more interested in whether or not there is some sort of definitive method of deducing the number of possible arrangements using a more formulaic method.

My guess is that a formulaic method would be complicated if we are trying to find one that works for all problems "of this type". It would be a good subject for mathematical research - if someone hasn't already tackled it.

To seek a formulaic method, we should first state explicitly what information is implicitly given in the problem. For example, the problem appears to say that the long side of the green block is equal to the long side of the yellow block plus the short side of the blue block. Try representing the sides of the blocks by variables and using the implied information to state relations among them.

If rectangle is formed by putting the blocks together, the length of one side of the rectangle is a sum formed by adding some of the variables and the length of other side is formed by a second sum made according to the rule: if a one dimension of a block is included in the first sum then include the other dimension in the second sum. If no dimension of a block is included in the first sum then include one of the dimensions of the block in the second sum.

That would be a combinatorial problem, but we would have to analyze each combination to see that it formed a complete rectangle instead of figure with a gap in it.

With more blocks, we'd have to worry that some configuration of blocks might be internal to the entire assembly, so the sums for the dimensions of the assembled rectangle would not directly involve the dimensions of the internal configurations.