Relative Minus signs from different Feynman Diagrams I have a problem understanding the occurrence of a the relative minus signs between contributions, coming from different Feynman diagrams, involving fermions. A simple example is Bhabha scattering $e^+ e^- \to e^+ e^-$. This process can happen by scattering or annihilation. I know the heuristic argument as mentioned for example here and in many books. I'm trying to understand this by computation using the S-Matrix expansion. 
Disclaimer: I will use quite sloppy notation to come as quickly as possible to my question.
We have $\lvert i\rangle = \lvert e^+ e^-\rangle =  c^\dagger d^\dagger \lvert 0\rangle$ and $\langle f\rvert = \langle 0\rvert dc$ 
The contribution part of the second order S Matrix term is for the scattering diagram (ignoring many things)
$$ S^a \propto (\bar \Psi^- \Psi^+)_{x_1}(\bar \Psi^+ \Psi^-)_{x_2}$$
and for the annihilation diagram
$$ S^a \propto (\bar \Psi^- \Psi^-)_{x_1}(\bar \Psi^+ \Psi^+)_{x_2}$$
The corresponding amplitudes are therefore (again focusing only on the sign relevant parts)
$$\langle f\rvert S^a\lvert i\rangle \propto \langle 0\rvert cd N\{ c^\dagger c dd^\dagger\} c^\dagger d^\dagger \lvert 0\rangle$$
and 
$$\langle f\rvert S^b\lvert i\rangle \propto \langle 0\rvert cd N\{ c^\dagger d^\dagger d c\} c^\dagger d^\dagger \lvert 0\rangle$$
I have read the corresponding pages in quite a few books, and the standard ways to explain the minus sign are:
I That we need now to bring both terms into equal normal order (Mandl-Shaw)
or
II that we need to make sure that a $c$ always stands next to an $c^\dagger$ and equally for $d$, i.e. make sure a particle is always annihilated after it is created before another particle is created. (See for example (Quantum Field Theory and the Standard Model - Schwartz)
Using the anti-commutation relations between the creation and annihilation operators leads for both demands to a relative minus sign between the two contributions. My problem is understanding where the need for I or II comes from? In other words: If I follow the instructions in the textbooks I get the correct result, which is the same as if I used the heuristic rule mentioned at the beginning. Anyway I do not understand where these rules comes from.
Why do we need to bring the operators in both amplitudes into equal normal order? Or
Why do we need to annihilate a particle as soon as it was created before another particle is created?
Any help or reading suggestion would be much appreciated
 A: First, the second equation starting with $S^a\propto\dots $  should probably say $S^b\propto\dots$.
Now, the first two equations for the operators $S^a$ and $S^b$ which are the relevant parts of $S=S^a+S^b$ have the positive plus sign – the additional factors that are omitted don't differ by any extra sign because there is a well-defined factor (and sign) in front of the $\bar\Psi\Psi$ factor of the interaction term in the Lagrangian. You omitted the coefficient $K_a$ and $K_b$ in the two equations, respectively (some photon propagators and other things).
Without the (correct) relative sign flip, the total amplitude would be proportional to $K_a+K_b$.
To proceed (and fix the sign), it is enough to notice that in the last two displayed equations,
$$ \begin{eqnarray} 
N\{c^\dagger c d d^\dagger\} &= c^\dagger d^\dagger c d \\
N\{c^\dagger d^\dagger  dc\} &= c^\dagger d^\dagger d c
\end{eqnarray} $$
where I was careful to perform an even number of fermionic fields' transpositions (the permutation of $d^\dagger$ through $c$ and $d$ in the first evaluation, or nothing in the second evaluation) so that the manipulation is independent of the question whether $(-1)$ includes the sign flips for the transpositions or not. But the last expression from $S^b$ is, because $dc=-cd$, equal to minus the first one (the $cd$ and $dc$ at the end is the only difference between the two), which is why, after the conversion of everything to the multiples of $c^\dagger d^\dagger cd $ which is still sandwiched in between the (fixed) initial and final states, produces an amplitude proportional to $K_a-K_b$, with the relative minus sign.
To answer "why I or II", I would choose "why I". The reason why we need to bring both terms into the same normal-ordered form is that we want to factorize the coefficients. But for $A_b=-A_a$ with the minus sign coming from the simple counting of permutations of the annihilation operators inside $S$ (or creation operators inside $S$), the distribution law is only possible if $A_a$ may be taken out of the parenthesis i.e. factorized, i.e. if we convert $A_b$ to $-A_a$ first:
$$ K_a \cdot A_a + K_b\cdot A_b = K_a \cdot A_a - K_b\cdot A_a = (K_a-K_b)\cdot A_a$$
