How is the dynamic equilibrium nature of fermi-dirac distribution of particles facilitated? I read this in Kittel: Introduction to Solid State Physics about deriving that product of electron and hole concentration as independent at a given temperature by the law of mass action. 
For this model, it is assumed wherein the electron-hole pairs are generated using photons of black body radiation at rate $A(T)$, with $T$ being temperature, within the semiconductor and the recombination rate of pair at $B(T)np$. At equilibrium
$\frac{dn}{dt} = A(T) - B(T)np=0 \implies np = \frac{A(T)}{B(T)}$
Is this model a specific case of facilitating dynamic equilibrium nature of Fermi-Dirac distribution or the principle way of how fermions interact through photons?
 A: Photons are modeled as bosons with an integer spin, have a symmetry and can occupy the same quantum state. 
All this means, is that they use the Bose-Einsten distribution instead. Where the Bose Einstein distribution gives the average number of Bosons found in an energy state, $\epsilon$.
$$\overline{n}=f_{-}{(\epsilon)}=\frac{1}{\exp(\beta(\epsilon-\mu)-1}$$
$$\mu=\left(\frac{\partial{F}}{\partial{N}}\right)_{T,V}\text{is the chemical potential.}$$
$$\beta = \frac{1}{k_{B}T}$$.
Boltzmann's constant, $k_{B}=1.38 \times 10^{-23}\mathrm{ m^2 kg s^{-2} K^{-1}}$
$T$ is the temperature in Kelvin. 
Remember fermions $\frac{1}{2}$ spin and are not symmetrical. They also obey the pauli exclusion principle. Does that sound like photons to you? Their average number is given by the Fermi-Dirac.
$$\overline{n}=f_{+}{(\epsilon)}=\frac{1}{\exp(\beta(\epsilon-\mu)+1}$$
You may already know that stuff but I am guessing you are confused at the blackbody model. The model for blackbody radiation is a cavity. 
In this situation energy is quantised such that $\epsilon = nh\upsilon$ with $h=6.64\times 10^{-34}\mathrm{m^2kgs^{-1}}$ which means that the mean number of particles in a given energy state is given by
$$\overline{n}=f_{-}{(\upsilon)}=\frac{1}{\exp(\beta(h\upsilon-\mu)-1}$$ 
