Did I get this BJT configurations correct?
Let's try to find some sense in this formal classification...

Transistor vs circuit input. The input of the bipolar transistor is its base-emitter junction; so there we have to apply the input voltage. In the general case, the base-emitter junction is floating... but we prefer to deal with single-ended (grounded) voltages. Then we present the floating base-emitter voltage as a difference between two single-ended input voltages. Note they are "circuit-input voltages" while the base-emitter voltage is "transistor-input voltage". You can see the same difference in op-amp amplifying circuits. For example, in an op-amp inverting amplifier, VIN is the circuit-input voltage... while V(+) - V(-) is the op-amp input voltage.

So, we can drive the transistor in a few ways giving the names in this classification:

Common-emitter stage. In this configuration, we fix the emitter voltage (by inserting another voltage source, Zener diode, capacitor... or simply grounding the emitter). Then we vary the base voltage (i.e., we drive the transistor from the side of the base). When the input base voltage increases, the collector current increases and the collector voltage decreases; so this is an inverting amplifier.

Common-base stage. Now we fix the base voltage and vary the emitter voltage (we drive the transistor from the side of the emitter). When the input emitter voltage increases, the base-emitter voltage decreases, the collector current decreases and the collector voltage increases; so this is a non-inverting amplifier.

Common-collector stage. Here the collector voltage is fixed at +VCC  (as though the collector is AC grounded); hence the name "common-collector". But this is a formal statement that has nothing to do with the transistor-input voltage; it only serves to somehow put another circuit in this classification. As above, the input voltage has to be applied between the base and emitter. So, like in the common-emitter stage, we apply the circuit-input voltage to the base (drive the transistor from the side of the base) but here the transistor produces the emitter voltage. It does it by passing its collector current through an emitter resistor - the so-called emitter degeneration). So, when the input base voltage increases, the collector current increases, the emitter voltage increases thus decreasing the base voltage and collector current. The result is the emitter voltage follows the base voltage; the circuit acts as a voltage follower.

Common-emitter stage with emitter degeneration. This is a combination of common-emitter and common-collector stages since resistors are inserted in both collector and emitter. This circuit is difficult to fit into the classification because, in practice, it has no common (fixed) terminal. If we shunt the emitter resistor by a capacitor, it will become an "AC common-emitter stage".

Differential stage. Finally, we can control the transistor-input voltage by changing simultaneously both circuit-input voltages. First we can do it in a differential manner changing them in opposite directions. The collector current and voltage will change more vigorously than in the case of the common-emitter and common-base stage.

Then, we can do it in a common-mode manner changing them in the same direction. Their difference across the base-emitter junction will not change; the collector current and voltage will not change as well.



I think that after these explanations, you should be able to correct your circuits so that they fit into the classification.