Counting matchings, the modern way A hundred years ago, if you had $k$ men and $k$ women and wanted to marry them all off in pairs, it was easy to see that there are exactly $k!$ ways to do that.
Today, however, societal standards have evolved, and many countries now recognize same-sex marriages. Mathematically we can model that by saying that men can now be brides and women can now be grooms. Each wedding still needs one bride and one groom, however, if only because one of the spouses needs to be listed in the left rubric on the marriage certificate and the other on the right.
How many ways are there now to marry off our $2k$ persons in pairs? A bit of elementary combinatorics and algebra shows that we have
$$ \tag{$\dagger$} \frac{(2k)!}{k!} = (k+1)\cdot(k+2)\cdot(k+3)\cdots(2k-2)\cdot (2k-1) \cdot 2k $$
possibilities, when we want to remember who's the bride and who's the groom in each couple.
However, every way I can think of to justify this formula depends on first counting a larger number of somethings, and then dividing out some factors to account for overcounting. (If we justify $(2k)!/k!$ as $\binom{2k}{k}k!$ the division is hidden inside the argument that $\binom nk=\frac{n}{k!(n-k)!}$).

Question: Is there an intuitive way to explain the result $\text{($\dagger$)}$ where each of the $k$ factors have a discrete meaning? Or, in other words, an explicit description of a bijection from the set
  $$ \{1,2,\ldots,k,k+1\}\times\{1,2,\ldots,k,k+1,k+2\}\times\cdots\times\{1,2,\ldots,2k\} $$
  to the set of possible matchings?

 A: Let's assume the individuals are named by natural numbers between $1$ and $2k$. Order the happy couples as $(b_1, g_1), (b_2, g_2), \ldots (b_k, g_k)$ with $b_1 < b_2 < \ldots < b_k$. Then you have $2k$ choices for $g_1$, $2k-1$ choices for $g_2$ and so on giving $2k(2k-1)\ldots (k+2)(k+1)$ possibilities for the $g_i$. When you know the $g_i$ the $b_i$ are determined by the ordering.
A: To flesh out my comment a bit: we can count the $(2k-1)!!$ unordered pairings simply: line the $2k$ people up and number them.  Now, there are $(2k-1)$ people that person 1 can be married to; there are $(2k-3)$ people that the next-smallest unmarried person can be married to; etc.  Once we have our couples, we can assign a 'preferred partner' to each one independently, giving $2^k$ choices (this can be expressed easily in boolean form as 'is the smaller or larger of our pair the preferred?').  Of course from here, showing that $2^k(2k-1)!!=\frac{(2k)!}{k!}$ is a little trickier - one can pair off factors, but that's a surprisingly painful process.
A: Put everyone in a row, choose the $k$ brides (with order) and then marry the $j$-th bride chosen to the $j$-th groom left standing.
