The question: "I'm wondering if there is some good reason why the universe as we know it has to have twelve particles rather than just four."
The short answer: Our current standard description of the spin-1/2 property of the elementary particles is incomplete. A more complete theory would require that these particles arrive in 3 generations.
The medium answer: The spin-1/2 of the elementary fermions is an emergent property. The more fundamental spin property acts like position in that the Heisenberg uncertainty principle applies to consecutive measurements of the fundamental spin the same way the HUP applies to position measurements. This fundamental spin is invisible to us because it is renormalized away. What's left is three generations of the particle, each with the usual spin-1/2.
When a particle moves through positions it does so by way of an interaction between position and momentum. These are complementary variables. The equivalent concept for spin-1/2 is "Mutually unbiased bases" or MUBs. There are only (at most) three MUBs for spin-1/2. Letting a particle's spin move among them means that the number of degrees of freedom of the particle have tripled. So when you find the long time propagators over that Hopf algebra you end up with three times the usual number of particles. Hence there are three generations.
The long answer: The two (more or less classical) things we can theoretically measure for a spin-1/2 particle are its position and its spin. If we measure its spin, the spin is then forced into an eigenstate of spin so that measuring it again gives the same result. That is, a measurement of spin causes the spin to be determined. On the other hand, if we measure its position, then by the Heisenberg uncertainty principle, we will cause an unknown change to its momentum. The change in momentum makes it impossible for us to predict the result of a subsequent position measurement.
As quantum physicists, we long ago grew accustomed to this bizarre behavior. But imagine that nature is parsimonious with her underlying machinery. If so, we'd expect the fundamental (i.e. before renormalization) measurements of a spin-1/2 particle's position and spin to be similar. For such a theory to work, one must show that after renormalization, one obtains the usual spin-1/2.
A possible solution to this conundrum is given in the paper:
Found.Phys.40:1681-1699,(2010), Carl Brannen, Spin Path Integrals and Generations
http://arxiv.org/abs/1006.3114
The paper is a straightforward QFT resummation calculation. It assumes a strange (to us) spin-1/2 where measurements act like the not so strange position measurements. It resums the propagators for the theory and finds that the strange behavior disappears over long times. The long time propagators are equivalent to the usual spin-1/2. Furthermore, they appear in three generations. And it shows that the long time propagators have a form that matches the mysterious lepton mass formulas of Yoshio Koide.
Peer review: The paper was peer-reviewed through an arduous process of three reviewers. As with any journal article it had a managing editor, and a chief editor. Complaints about the physics have already been made by competent physicists who took the trouble of carefully reading the paper. It's unlikely that someone making a quick read of the paper is going to find something that hasn't already been argued through. The paper was selected by the chief editor of Found. Phys. as suitable for publication in that journal and so published last year.
The chief editor of Found. Phys. is now Gerard 't Hooft. His attitude on publishing junk is quite clear, he writes
How to become a bad theoretical physicist
On your way towards becoming a bad
theoretician, take your own immature
theory, stop checking it for mistakes,
don't listen to colleagues who do spot
weaknesses, and start admiring your
own infallible intelligence. Try to
overshout all your critics, and have
your work published anyway. If the
well-established science media refuse
to publish your work, start your own
publishing company and edit your own
books. If you are really clever you
can find yourself a formerly
professional physics journal where the
chief editor is asleep.
http://www.phys.uu.nl/~thooft/theoristbad.html
One hopes that 't Hooft wasn't asleep when he allowed this paper to be published.
Extensions: My next paper on the subject extends the above calculation to obtain the weak hypercharge and weak isospin quantum numbers. It uses methods similar to the above, that is, the calculation of long time propagators, but uses a more sophisticated method of manipulating the Feynman diagrams called "Hopf algebra" or "quantum algebra". I'm figuring on sending it in to the same journal. It's close to getting finished, I basically need to reread it over and over and add references:
http://brannenworks.com/E8/HopfWeakQNs.pdf
The most important difference between strange and charm quarks is their mass.
The stable u-quark has mass of only 2.7 MeV. The mass of the s-quark is about 95 MeV. In contrast the mass of the c-quark is 1275 MeV. Because $m_c-m_s$ is much larger than $m_s-m_u$. Thus the phase space of decay products is much larger and probability is amplified.
However there's extra reason for s-quark decay to be slow. The dominant decay of the c-quark is to the s-quark. If you know that the weak interaction is governed by SU(2) gauge field with each quark generation forming a doublet under this gauge group, you would expect that W-boson will convert upper quark like c-quark into its partner from the same generation i.e. s-quark. However s-quark has to decay into u-quark from the different generation. This is possible because mass eigenstates with which we actually define generations are not the same as interaction eigenstates. W-boson thus treats s-quark as having a small admixture of d-quark. This admixture is characterized by the Cabbibo angle with $\sin\theta_c\approx 0.22$. Thanks to it being nonzero s-quark decays but as it's not very large the decay goes slowly.
Best Answer
In particle physics we have a list of "quantum numbers" that describe a particle. Different types of interactions may conserve, or may not conserve, different quantum numbers.
You give the example of decays which change quark flavor. Baryons and mesons (which we model as being made of quarks, even though individual quarks are confined) are assigned "flavor quantum numbers": the $D$ mesons have charm quantum number $C=\pm1$, the $K$ mesons have strangeness $S=\pm1$ but charm $C=0$, and so on. The strong and electromagnetic interactions do not change the flavor quantum numbers in a system, but the weak interaction does. So the stoichiometry skills that you learned in chemistry work for strong-interaction scattering, such as the strangeness-conserving production of hypernuclei, but not for weak decays which change those quantum numbers.
(In fact, you might say that it only makes sense for us to talk about flavor quantum numbers because the interaction that changes them is weak.)
A few of these quantum numbers are conserved by all known interactions. Those include
When you do chemical stoichiometry, like in your calcium carbonate decomposition reaction, you're conserving electric charge, the number of electrons, and the number of protons and neutrons. Your baryon number conservation is constrained because there's no interaction at the energies chemists care about which allows protons to change into neutrons or vice-versa, so you have to conserve proton and neutron numbers separately. Furthermore there's no interaction, at the energies that chemists care about, which allows a nucleon to hop from one nucleus to another, so you have to separately conserve the number of calciums, the number of carbons, etc.
It's tempting and useful to take these conservation laws and use them to conclude that a calcium nucleus is "made of" twenty protons and twenty-ish neutrons. But that approach breaks down when you start to consider the flavor-changing weak interactions. The muon decays by the weak interaction into a neutrino, an antineutrino, and an electron; but there's evidence against any model where the muon "contains" those decay products in the way that we can say a nucleus "contains" nucleons.