Why does carbon-12 decay take so much longer to occur than alpha decay? In this specific question, I'm talking about why there is such a large difference in the time taken to emit a carbon nucleus, than an alpha nucleus. 
In a recent lecture, my professor discussed the fact that heavy elements can emit a carbon nucleus as well as an alpha nucleus, but I missed out on the part when he discussed the reasons why we don't see carbon decay very often/it takes such a long time. 
Any clarification on this would be appreciated.
 A: The decay rate for alpha emission can be approximated by modelling the process as due to the particle quantum-tunneling out of a potential. The logarithm of the half-life is an additive constant (dependent on how we nondimensionalise time) plus $2c\pi\alpha Z_1 (Z-Z_1)\sqrt{2m/E}$, where the emitted particle has mass $m$, kinetic energy $E$ and nucleon number $Z_1$, and $Z$ is the nucleon number of the parent nucleus. While $Z-Z_1$ switches from $Z-4$ for alpha decay to $Z-12$ for carbon-12 decay, which is a little smaller than $Z-4$ for the kind of large nuclei susceptible to carbon-12 decay, the $Z_1\sqrt{m}$ factor is proportional to $Z_1^{3/2}$ so is $5.2$ times larger in the carbon-12 decay. The log half-life is thus somewhat larger in this case, making the half-life itself much longer than for alpha emission.
A: In  https://en.wikipedia.org/wiki/Cluster_decay - pointed out by @JonCuster one finds 
$\lambda = \ln 2 /T_c = \nu S P$
Comparing with $\alpha$ decay, $S$ - formation probability for heavier fragments can be much lower  than $\alpha$, because they have more complicated structure, while $\alpha$ is a very good cluster in nucleus in many cases (in Li, C, ...).
Another term is $P$ - penetrability, which translates to possibility to overcome a Coulomb barrier. What is the "penetrability factor"? - but from elsewhere you find this equation:
$P \sim \exp(-2\pi z_1 z_2 e^2/\hbar v)$
From this one can see that (if you consider Z=92 nucleus) $z_1z_2$ is 180 for $\alpha$ (2*90) and 516 for carbon (6*86). Then you must have higher velocity $v$ for a similar $P$, which means a lower fraction of the the wavefunction can participate. And exponential makes it even more important.
One is used to a fact, that the Coulomb barrier plays a role when particle is approaching the nucleus from outside, but it is actually similar when the particle resides inside, because it is low in a potential well created by some field of all particles.
A: A simple (if hand-wavy) answer is that one part of the wavefunction for a carbon nucleus is a cluster of three alpha particles.  Therefore, while you're waiting around for your carbon nucleus to form so that it can possibly tunnel out to cause a cluster decay, you're necessarily forming alpha particles that are more likely to tunnel out themselves than to cluster together inside the nucleus.
