So, here's the problem:

Given $n$ distinct sets $X_{1}, X_{2}, ... X_{n}$, find all possible sets consisting of $n$ elements, one from each set. Such a result is known as the Cartesian product, and can be represented mathematically as:

$P = \prod_{j=1}^{n}X_{j} = X_{1}\times X_{2}\times ...X_{n}$

where $P$ is the actual product. What does this product look like? Let's see an example.

We'll say we have two sets $X_{1}$ and $X_{2}$:

$X_{1} = \{a, b\}$
$X_{2} = \{1, 2\}$

The Cartesian product of these two sets would be a "set of sets", or more formally a collection or family of sets, and would look like the following:

$P = X_{1}\times X_{2} = \{\{a,1\},\{a,2\},\{b,1\},\{b,2\}\}$

As you can see if you examine the above, for each element of set $X_{1}$, there are exactly two sets in the result that contain that element - each of which also contains one or the other element of $X_{2}$. We have our product. Now let's figure out how we got here - and we'll start with the easy-to-compute question: How many possible combinations can there be?

Well, this is likely the easiest problem in combinatorics there is to solve. Remember above where we stated the formula for arriving at a Cartesian product?

$P = \prod_{j=1}^{n}X_{j} = X_{1}\times X_{2}\times ...X_{n}$

Well, the number we're particularly after is the cardinality of the final collection $P$. In this context, the word "cardinality" means "size" - i.e. the cardinality of any set is the number of elements in that set. In mathematical notation, we represent the cardinality of a set by surrounding the name of the set by two vertical bars - very similar to what you may know as the notation for absolute value. So, the cardinality of the set $X_{1}$ above would be represented as $\left|X_{1}\right|$ and would be equal to 2, because set $X_{1}$ has exactly 2 elements, $a$ and $b$. So, given what we now know about cardinality, how do we use that to compute the cardinality of $P$, or $\left|P\right|$? It literally couldn't be easier:

$\left|P\right| = \prod_{j=1}^{n}\left|X_{j}\right| = \left|X_{1}\right|\times \left|X_{2}\right|\times ...\left|X_{n}\right|$

That's right - you just multiply the cardinalities of all the sets in question together, and voilá: you have how many combinations there would be. Now, we've already stated that the cardinalty $\left|X_{1}\right|$ of set $X_{1}$ is equal to two, and it should be plain by now that the cardinality $\left|X_{2}\right|$ of set $X_{2}$ is also equal to two. As such, that would give us the answer 4 for $\left|P\right|$, and lo and behold, the result set $P$ shown above has exactly 4 elements!

Now to the harder bit - how do we figure out precisely what those elements are? Enter my good friend and soon-to-be yours (if he isn't already), recursion. Below I have implemented the algorithm I worked out - though I make no claims to it being a heretofore undiscovered one - as a pure function (i.e. a function "without side-effects", which does not alter any external state) in Ruby, my programming language of choice. Note how the method calls itself as part of its execution - that's recursion.

def cartesian_product(*enums)
  return Set.new([Set.new]) if enums.empty?
  combos = Set.new
  first = enums.shift
  subproducts = cartesian_product(*enums)
  subproducts.each do |product|
    first.each do |element|
      combos.add(product.clone.add(element))
    end
  end
  combos
end

So there's the meat of the matter - more explanation to come in part 2, coming soon!

That's science - so stay tuned.