Surprising limit (probability of no two coinciding pairs) I stumbled upon this question by random chance. The motivation is kind of long, the question is pretty short; if you're just here for the limits, feel free to skip to the break.
I'm taking five courses this semester, all of which meet on two days, and I happened to notice that all of my days seemed a little bit different. So I took a closer look at my schedule, and sure enough, none of my classes share the same two days.
This struck me as very unlikely, so I decided to work out the probability of this occurring. Of course, in reality, there are all sorts of complicating factors that make certain distributions more favorable than others, but I decided to ignore that and just assume that the schedules were independent identically distributed variables.
It was sort of convenient that there are also five days in a week, so the motivating question was this: In an $n$ day week, what is the probability that $n$ classes meeting twice each, do not have any pair that meets on the same two days?
So I looked at the combinatorics, and I'm pretty sure the answer is the obnoxiously vertical formula
$$\frac{ \displaystyle  { {n \choose 2} \choose n} n!}{ {n \choose 2}^n}$$
The idea here is that there are ${n \choose 2}$ possible schedules, so the list of all possible outcomes includes $n$ choices with repetition from this pool, considering the classes distinguishable. The list of desired outcomes chooses $n$ choices without repetition from this pool, and then assigns the schedules to the classes.
[ Perhaps this is not right; if not I'd be interested in that, too, but my question for you all is really about this expression, not my class situation :) ]
Now I knew by working out by hand that $n=3$ gives about 22% and $n=4$ gives about 28%, and I computed my actual situation at about 30%. Monotone increasing is what I expected. After all, you have dramatically more ability to separate the classes during longer weeks. 
I also expected that this effect would quickly dominate the complication that more classes are being added in. But the picture that arises for (larger) small $n$ made it pretty clear that I was wrong. 

When I was playing around in WolframAlpha, I found the limit
$$\lim_{n\to\infty}\frac{ \displaystyle  { {n \choose 2} \choose n} n!}{ {n \choose 2}^n} = \frac{1}{e}$$
That caught me off guard. I immediately thought of Stirling's approximation, but I can't make it fit, since in Stirling $e$ is raised to the $n^\text{th}$. Admittedly I don't know a lot of special limits, but writing it out in terms of factorials seems to get nowhere near the definition of $e$. 
Is there an elementary explanation for this limit?
Bonus question: "choose $2$" seems to be a pretty special circumstance: if the twos are replaced by ones, the limit becomes $0$, and if the twos are replaced by threes or fours, the limit becomes $1$. My guess is that with the right explanation of the interesting case, the reason for these will become pretty obvious.
 A: The situation is the generalised Brithday problem. You have $m$ possible choices (schedules in your setup), and pick $n$ items with possible repetitions. What is the probability that you never pick the same item twice?
The probability is, for independent choices,
$$\frac{\binom{m}{n}n!}{m^n} = \frac{m!}{m^n(m-n)!} = \prod_{k=0}^{n-1}\frac{m-k}{m} = \prod_{k=1}^{n-1}\left(1 - \frac{k}{m}\right).$$
When $k \ll m$, we can approximate that well using the first order approximation of $\log (1-x)$,
$$\begin{align}
\prod_{k=1}^{n-1} \left(1 - \frac{k}{m}\right) &= \exp \left(\sum_{k=1}^{n-1}\log \left(1-\frac{k}{m}\right) \right)\\
&\approx \exp \left(-\sum_{k=1}^{n-1}\frac{k}{m} \right)\\
&= \exp \left(-\frac{n(n-1)}{2m} \right).
\end{align}$$
In your case, you have $m = \binom{n}{2}$ schedules, so the argument of the exponential function comes out to be exactly $-1$.
We can - still for $n \ll m$ - use more terms of the expansion of the logarithms to get a more precise approximation,
$$\log \left(1 - \frac{k}{m}\right) = - \sum_{j=1}^\infty \frac{1}{j}\left(\frac{k}{m}\right)^j = - \left(\frac{k}{m} + \frac{k^2}{2m^2}\right) + O\left(\frac{k^3}{m^3}\right).$$
That yields
$$\prod_{k=1}^{n-1}\left(1-\frac{k}{m}\right) = \exp\left(- \frac{n(n-1)}{2m} - \frac{(n-1)n(2n-1)}{12m^2} + O\left(\frac{n^4}{m^3}\right)\right)$$
and inserting $m = \binom{n}{2} = \frac{n(n-1)}{2}$
$$\begin{align}
\prod_{k=1}^{n-1}\left(1-\frac{k}{m}\right) &= \exp\left(-1 -\frac{2n-1}{2n(n-1)} + O(n^{-2})\right)\\
&= \frac1e\exp\left(-\frac{1}{n} + O(n^{-2})\right)\\
&= \frac1e\frac{n-1}{n} + O(n^{-2}).
\end{align}$$
