If you're familiar with double-subscript notation, you have your answer at hand.
For example, the base-emitter voltage \$v_{BE} \$ is positive when the base is more positive than the emitter, i.e., for \$v_{BE}\$, the \$ +\$ sign is at the base node.
Likewise, the emitter-base voltage \$v_{EB} \$ is positive when the emitter is more positive than the base, i.e., for \$v_{EB}\$, the \$ +\$ sign is at the emitter node.
With that in mind, for NPN transistors, the equations are written in terms of \$v_{BE}, v_{CB}, v_{CE}\$. By KVL, \$v_{BE} + v_{CB} = v_{CE}\$ so, if you know any two, you know the third.
Now, remembering the structure of NPN transistor, it is the case that the base-emitter junction is forward biased when \$v_{BE}\$ is positive and the base-collector junction is reverse biased when \$v_{CB}\$ is positive.
The cutoff region is formally defined as the condition that both junctions are reverse biased: \$v_{BE}< 0, v_{CB}>0\$
For PNP transistors, simply reverse the order of the subscripts and everything follows through.
The ULN2003A is a Darlington array, made up of Darlington pairs. The Darlington pair has higher current gain than an individual bipolar transistor. The side effects include a higher \$V_{be}\$ (in common emitter configuration, it will appear as about two diode drops instead of one) and \$V_{ce}\$ (\$V_{be}\$ of one transistor, plus \$V_{ce}\$ of the other).
Best Answer
First, you mix voltage and current. Voltage is the "pressure" on electrical charges, so they want to move. Current is the flow of electrical charges, if the find a way.
Case 1:
To answer the question, have a look into this datasheet of a common BC337
The table on page 2 states a Collector Cutoff Current of 100nA for V_CE=45V and V_BE=0V.
So yes, there will be a current through the collector. Keep in mind that a NPN transistor looks like two diodes connected back to back:
simulate this circuit – Schematic created using CircuitLab
While this doesn't describe a transistor in general, here it does. You are operating the upper diode in reverse direction. In this case, real diodes will show a small leakage current, 100nA is a perfect sample for it.
You didn't say where E should be connected to, but if it's connected to GND, as in my schematic, the lower diode is shorted by the connection between B and GND, so the 100nA will flow through the base leg. It's just the same as if E is not connected anywhere. If E is also connected to Vcc, you just get a second diode in parallel to the first, so about twice the current (depending of the characteristics of the diodes)
Case 2
Well, if only B is connected to Vcc, and the rest is not connected to anything, no current will flow. If you connect E and/or C to GND, the diodes will be operated in forward direction, and a significant current will flow (and destroy the transistor if currents are niot limited by resistors)