Why there aren't classical field associated with fermions? In A Introduction to Nuclear Physics by GreenWood, It's written

Bosons are particles that obey Bose-Einstein statistics and are
characterized by the property that any number of particles may be
assigned the same single-particle wave function. Thus, in the case of
bosons, coherent waves of macroscopic amplitude can be constructed,
and such waves may to a good approximation be described classically. For example, photons are bosons and the corresponding classical field is the familiar electromagnetic field $\mathbf{E}$ and $\mathbf{B}$, which satisfies Maxwell's equation.

I don't understand the italic part.
Later on the author says that Fermions interact through the fields of which they are sources. The particle associated with the interaction fields are bosons.
The first passage says that bosons can be approximated to wave while the second says that fermions produce fields. Isn't it should be that fermions produce bosons that can be approximated to wave?
I don't understand How this all follows from the Symmetric and anti-symmetric wave function. Is there any simpler way to understand this without getting into much detail (that is without getting the detail of QFT)?
 A: You probably heard of the quantum mechanical description of a wave function of a subatomic particle: It's probability to be measured at a certain point in space follows a wave like structure. If you are not familiar with that look at the double-slit experiment. This is why, in the earliest versions of quantum mechanics, subatomic particles are described as wave functions. The easiest interpretation of the wave function is that the (space-dependent) value of its square is proportional to the probability (density) to observe the particle at a certain point (You actually have to integrate the square of the wave function over a finite region in space to find the probability to find the particle in this region). While the interpretation of this is still debated, in case of ONE electron this wave function ceases to exist as soon as this one particle has been detected somewhere.
If you now think of electromagnetic waves, for example waves that a radio can pick up, you can observe, more easily for long wavelenghts, that the probability to pick up signal, i.e. detect the wave, also follows a wave like structure. (Hence, obviously, the name electromagnetic wave). Compared to the electron I talked about before, the wave does not cease to exist as soon as you pick up the signal once, since it, quantum field theoretically spoken, consits of a really (!) high number of photons. In such an electromagnetic wave, like the one in a microwave oven, or a radio signal, all the photons have the same wave function (ignoring modulation of the signal to actually decode sound waves or other modulations of the signal), so you can detect a lot of photons everywhere, each of which you can describe by the same probability distribution. Therefore you can model this wave function as classical wave. There are just so many photons, that most of the effects we observe become continuous in space and time.
Why is there now such a difference between photons and electrons? Photons are bosons, and this means by definition that many photons can share the same wave function (or more correct, the same quantum state). Electrons are fermions and that means by definition that they cannot share the same wave function. If you do not want to go into details of QFT, you have to accept this difference just as given.
How does it relate to symmetric and anti-symmetric wave functions?
Bosons have symmetric wave functions, in a sense that exchanging two of the particles making up the combined wave function does not change the wave function.
Ignoring all other possible quantum numbers of the particles (spin, etc.) and just looking at the position of the wave function, this symmetry of the wave function $\Psi$ can be expressed  for two bosons as $\Psi(x_i,x_j)=\Psi(x_j,x_i)$. Exchanging the positions of the two particles does not change the wave function. This means if $x_i=x_j$, both particles "at the same place", we just have $\Psi(x_i,x_i)=\Psi(x_i,x_i)$, which is trivially fulfilled for all wave functions (I mention that this is, of course, not the only condition to find a valid wave function).
In case of fermions, again only caring for position of the particle and ignoring all other quantum numbers, the anti-symmetric condition can be expressed as $\Psi(x_i,x_j)=-\Psi(x_j,x_i)$. Exchanging the particles brings a minus sign. If we now set $x_i=x_j$ we have as condition $\Psi(x_i,x_i)=-\Psi(x_i,x_i)$, and this can only be fulfilled when $\Psi(x_i,x_i)=0$. Therefore we conclude that two fermions (and certainly also not more than two) with the same quantum numbers can be at the same position (where you should keep in mind that here with position I mean the spatially extended wave function).
This concludes that we cannot have classically described "fermionic waves", where myriads of fermions have the same wave function and behave almost like a classical wave, like it is possible with photons.
I am sorry, but I do not understand the question about how "fermions produce bosons that can be approximated to wave?".
