systems of particles that are not symmetric or anti-symmetric; Helium 4 Suppose I have an electron and a proton, and that the electron is in the spin-up state, and that the proton is in the spin-down state. The particles are distinguishable, so I should just be able to write the wavefunction as 
$$|\Psi_{1}\rangle=|\uparrow\rangle_{e}|\downarrow\rangle_{p}$$
Clearly, this wavefunction is neither symmetric nor antisymmetric, since $|\Psi_1\rangle$ is not at all proportional to 
$$|\Psi_{2}\rangle=|\uparrow\rangle_{p}|\downarrow\rangle_{e}$$
So, from that standpoint this would seem to be neither a fermion or a boson. However, the particle also wouldn't be an angular momentum / spin eigenfunction; it would be in an even superposition of one of the spin-1 triplet states and the spin-0 singlet state. So, if you measured it's total spin, it would be integral, which seems to indicate that it would still be a boson, even though the wavefunction is not symmetric under particle exchange. Which is it? Is there another name for it? What type of statistics would such particles follow then?
Along this line of thought, it seems strange to me that a larger system of particles, such as helium 4, or heavier elements, being large combinations of protons, neutrons, and fermions, should conspire to create a perfectly symmetric or anti-symmetric wavefunction as required for the behavior we see of such particles, since such combinations of wavefunctions form such an infinitely small fraction of all the other ways you could combine the wavefunction.
Can anyone help me?
 A: The question you have to ask, is what happens under swapping the entire sets of coordinates of two of the composite particles! You don't have to think of spin addition and the spin-statistics-theorem (especially since spin and orbital angular momentum cannot necessarly be kept apart for composite particles due to spin orbit coupling!).
Taking your hydrogen example, the wave function of two hydrogen atoms has a position and spin dependence for each electron:
$$\Psi(\vec r_{e1}, \sigma_{e1}, \vec r_{p1}, \sigma_{p1}, \vec r_{e2}, \sigma_{e2}, \vec r_{p2}, \sigma_{p2}) = -\Psi(\vec r_{e2}, \sigma_{e2}, \vec r_{p1}, \sigma_{p1}, \vec r_{e1}, \sigma_{e1}, \vec r_{p2}, \sigma_{p2}) = \Psi(\vec r_{e2}, \sigma_{e2}, \vec r_{p2}, \sigma_{p2}, \vec r_{e1}, \sigma_{e1}, \vec r_{p1}, \sigma_{p1})$$
Where the sign flips in each step as the electron and protons are both fermions.
The result of this procedure is the wave function where the position of both entire hydrogen atoms have been swapped. As there is no sign change, the wave function is symmetric in the coordinate sets for hydrogen, giving that hydrogen is a boson.
Following this procedure, you can easily see, that a system composed of an even number of fermions is a boson, and a system composed of an odd number of fermions must be a fermion.
