Wick's theorem: 4 field correlator, 2 different fields

[binbagsss]

Homework Statement

$T(\phi_1\Phi_2\phi_3\Phi_4)$

where $\phi_1$ is $\phi(x_1)$ and $\phi$ and $\Phi$ are two different fields.

By Wicks theorem $T(\phi_1\Phi_2\phi_3\Phi_4)= : : + contracted terms.$

QUESTION
Are the fully contracted terms (apologies for the bad notation I'm going to denote a contraction simply by a dot ).
a) $(\phi_1 . \phi_2)( \Phi_3. \Phi_4)$
b) $(\phi_1 . \phi_2)( \Phi_3 \Phi_4) + (\phi_1 \phi_2)( \Phi_3. \Phi_4) + (\phi_1 . \phi_2)( \Phi_3. \Phi_4)$

Homework Equations

look up, look down...

(yourtrousersarestillontodaybecauseihaventbeenonthevodka)

The Attempt at a Solution

to be honest I thought the answer would be b).
Whilst I understand that you can not contract between fields of different types, b) is consistent and does not do this, rather the other two fields (uncontracted are simply multiplying it.)

For example for a three field correlator of the SAME type I get $\phi_x G(y-z) + \phi_y G(x-z) + \phi_z G(x-y)$ (this will ofc vanish once sandwiched between the vacuum states, but as far as Wicks theorem takes us is this).

So in the same way I thought you could get contraction of the same field multiplied by uncontracted other fields, however having wrote out the whole expression, multiplied it all out in terms of creation and annihilation operators and applied the commutator relation etc I get a).

 

[binbagsss]

anyone?
I mean pretty much the definition of wicks thm before sandwiching between the vacuum states is correlator=normal order + ALL POSSIBLE contractions
Which reduces to, once sandwiched between the states, correlator=normal order + fully contracted

ta