There are 39 students in our class. We form groups of 2, with one left out. How many ways can the students be paired up? I know that the answer is $\frac{39!}{(2!)^{19}\cdot19!}$, where each pair can be organized $2!$ ways and the pairs can be arranged in $19!$ ways. We can also extrapolate the case for $5$ students, where I see that there are $15$ possible subsets ($\frac{5!}{2!^2\cdot2!}$). 
What I don't understand is why we don't account for where the single student can be placed in our solution. There are $19!$ ways to organize the pairs, but aren't there $20!$ ways to arrange the pairs and the one extra student?

Consider this alternate process:
First, arrange the students in some arbitrary order:  age, weight, IQ, whatever...
Now take a set of 39 cards.  Two are labelled "Group #1", two are labelled "Group #2", and so on, up to "Group #19".  The last, 39th card is labelled "On Your Own".  (Note we're now accounting for the lone student).  You are going to shuffle the cards, and then distribute them one by one, off the top, to the students in their arbitrary order.
How many distinct ways can you arrange 39 items, when there are 19 identical pairs and one item distinct from all the others? 
