What are the probability that the first two rows of the class are full? I was boring in my class. So I ask myself the question:

What are the probability that the first two rows of the class are full?

Knowing that we're $25$ students in my class and the class have $64$ seats ($8$ rows of $8$ seats).

$${8 \choose 8}{8 \choose 8}{48 \choose 9} = 1\ 677\ 106\ 640 \text{ ways to fill the first two rows}$$
and...
$${64 \choose 25} = 401\ 038\ 568\ 751\ 465\ 792 \text{ ways to fill the class}$$
thus,
$$\frac{1\ 677\ 106\ 640}{401\ 038\ 568\ 751\ 465\ 792} = 4,2 \times 10^{-7}\% \approx 0\%$$
Am I right? Because this seems impossible!
 A: That's not surprising at all, actually.  (Assuming, of course, that the students are seated randomly.)
To count the combinations, pick the $16$ students who will sit in the first two rows: ${25 \choose 16}$.  Pick the order that those students will fill the seats: $16!$.  For the remaining $9$ students, pick the set of seats they go in ${48 \choose 9}$, and then choose the order you seat them ($9!$) from front left to back right.
For the total number of cases, pick the set of seats ${64 \choose 25}$ and pick the order that you seat them in front left to back right ($25!$).
So yes, the ratio is a tiny number.
A great analogy to this comes from statistical mechanics.  Pretend the students are gas molecules.  What you're asking to do is spontaneously pressurize the front of the class, and create a vacuum in the back of the class.  This doesn't happen spontaneously, of course -- and the very reason is because the probability of this happening is so small!
That's exactly what you're seeing.
