So I'm trying to figure out the behavior of this system: you have $N$ coins, and every step, you choose one of the coins randomly and flip them.
Now we imagine a bazillion of these systems. We call $\rho_n$ the percentage of the systems that have $n$ coins flipped up (heads) -- state $S_n$.
It's easy to see that the number of systems that end up in $S_1$, for example, is all of the systems that were in $S_0$ (zero heads up, the only way to go is more heads), and $2/N$ of the systems in state $S_2$. With a little more of a stretch, it can be seen that the number of systems that end up in $S_2$ is $(N-1)/N$ of the systems that were in state $S_1$ and $3/N$ of the systems that were in state $S_3$. And so on, and so on.
We can then see that $\vec{\rho}'$ (the distribution of states after an iteration) is a simple matrix multiplication/linear transformation if $\vec{\rho}$, with the coefficients of the matrix being the ones listed above.
For example, for the case of $N = 3$, we have:
$$ \vec{\rho}' = \left[ \begin{array}{cccc} 0 & \frac{1}{3} & 0 & 0 \\ 1 & 0 & \frac{2}{3} & 0 \\ 0 & \frac{2}{3} & 0 & 1 \\ 0 & 0 & \frac{1}{3} & 0 \end{array} \right] \vec{\rho} $$
Which means that $\rho'_0$ (the new percentage of systems in state $S_0$) $= \frac{1}{3} \rho_1$ (1/3rds the percentage of systems that were in state $S_1$), that $\rho'_1 = \rho_0 + \frac{2}{3} \rho_2$, that $\rho'_2 = \frac{2}{3} \rho_1 + \rho_3$, etc.
More generally, for arbitrary $N$, the matrix is
$$ \left[ \begin{array}{cccccc} 0 & \frac{1}{N} & 0 & 0 & \cdots & 0 & 0 \\ 1 & 0 & \frac{2}{N} & 0 & \cdots & 0 & 0 \\ 0 & \frac{N-1}{N} & 0 & \frac{3}{N} & \cdots & 0 & 0 \\ 0 & 0 & \frac{N-2}{N} & 0 & \ddots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \frac{N-1}{N} & 0 \\ 0 & 0 & 0 & 0 & \frac{2}{N} & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & \frac{1}{N} & 0 \\ \end{array} \right] $$
I tried to find any steady states -- that is, distributions of $S_n$ that would remain unchanged under one transition. For something like $N=3$, you would expect something that "bulges" in the middle, kinda.
It turns out that the eigenvector corresponding to a steady state for $N=3$ is
$$ \vec{\rho} = \left[ \begin{array}{c} 1 \\ 3 \\ 3 \\ 1 \\ \end{array} \right] $$
Which are third row of the binomial coefficients (the third row of Pascal's Triangle).
This kind of makes a lot of intuitive sense --- you want something that peaks in the middle, and tapers off, kind of like a normal distribution. One could think of the rows of Pascal's Triangle as a sort of discrete normal distribution, so this is sort of understandable.
After testing out $N=2$, which is $[ \begin{array}{ccc} 1 & 2 & 1 \end{array} ]^T$, it appears that the eigenvectors corresponding to steady states of this transition are successive rows of Pascal's Triagngle, or the binomial coefficients.
Now I understand how the binomial coefficients can show up in something like an unbiased random walk. But this isn't an unbiased random walk --- the transition probabilities depend on the current state.
How do they show up here?
It's easy to verify that the vector $b$ of binomial coefficients is an eigenvector to the eigenvalue $1$, apart from the components $0$ and $N$ (which are easily verified separately), we have the equation
$$\begin{align} (A\cdot b)_k &= \frac{N+1-k}{N}\cdot b_{k-1} + \frac{k+1}{N}\cdot b-{k+1}\\ &= \frac{N+1-k}{N}\binom{N}{k-1} + \frac{k+1}{N}\binom{N}{k+1}\\ &= \binom{N-1}{k-1} + \binom{N-1}{k}\\ &= \binom{N}{k}. \end{align}$$
As to why $b$ is an eigenvector to the eigenvalue $1$, consider each coin separately. It is heads-up with probability $\frac12$, and tails-up with probability $\frac12$ (yes, I'm waving hands here). There are $\binom{N}{k}$ configurations with $k$ coins flipped heads-up.