Equilibrium constant. Can it be reached?
Dave's mathematical answer is perfectly valid, but I feel like another, perhaps more chemical point of view is based on the the law of large numbers, i.e., the "thermodynamic" limit.

"Equilibrium" is a statistical concept that only applies to large numbers of molecules.  Another way to say it is that equilibrium is a thermodynamic concept and thus only applies in the thermodynamic limit.  What is "large" depends on the particular system of interest.  In particular it depends on the equilibrium constant.  For your reaction of interest, $K_{eq} = 10$.  Thus the amount of $A$ and/or $B$ at equilibrium will be smaller than the amount of $C$.  For the equilibrium concept to apply to a system of $A$, $B$, and $C$ molecules, the calculated equilibrium amount of $A$ or $B$ molecules must be very large, say tens of thousands or so for precise work, or perhaps as low as hundreds for the equilibrium concept to be "approximately" correct.  

If the equilibrium constant were enormously large, say $K_{eq} = 10^{100}$, then by the same logic, the system would need to contain on the order of $10^{102}$ total molecules for the equilibrium to be meaningful.  (And by the way that is larger than the number of particles in the universe!)

Speaking loosely, if $K_{eq}$ is $10^{100}$ and there aren't close to that many molecules, the reaction is irreversible.

A probabilistic view of equilibrium for small numbers of particles, such as that mentioned in DavePhD's answer, is occasionally useful but not always valid.  For example, consider if a second reaction consumed $A$ irreversibly, converting it to $D$ at some rate.  If you take DavePhD's probabilistic view of the $\ce{A + B &lt;=&gt; C}$ equilibrium, you would assume that the probability distribution of persistence times for $A$ is exactly the same as in the thermodynamic limit.  However, that is not necessarily true with small systems.  Statistical fluctuations in the $\ce{A + B &lt;=&gt; C}$ system will mean that some of the time, $A$ persists far longer than in the thermodynamic case because due to a fluctuation, more than the "expected" amount of the molecules were in the $C$ form.  Some other times, $A$ will resist degradation to $D$ for far shorter than the thermodynamic equilibrium calculation would suggest.  To treat these cases, more complex equations such as the chemical master equation are required.