Collector current and Vcc in saturated BJT
In very simplified terms:


A saturated BJT's collector acts-like a voltage source between emitter and collector with only a small, but relatively constant voltage where \$10\:\text{mV} \le \quad\mid V_\text{CE}\mid\quad\le 1\:\text{V}\$. The depth of saturation varies from very light (when \$400\:\text{mV} \le \quad\mid V_\text{CE}\mid\quad\le 1\:\text{V}\$) to deep (when \$\mid\, V_\text{CE}\mid\:\lt 400\:\text{mV}\$.) The definition of saturation is usually taken to mean the situation when the collector voltage is driven closer to the emitter voltage than the base voltage is at relative to the emitter voltage (when the collector voltage is between the base and emitter voltages, in short.)
An active-mode BJT's collector acts-like a current source with a current determined by the relatively constant ratio, \$\beta=\frac{I_\text{C}}{I_\text{B}}\$, with the collector current set by the available base recombination current as \$I_\text{C}=\beta\cdot I_\text{B}\$ and independent of the collector load resistor value. A BJT is in active mode when its collector voltage isn't between its base and emitter voltages and is, instead, further away from the emitter voltage than it is from the base voltage.


The above are in broad strokes. If more pedantic about it, I'd have to spend time discussing magnitude vs sign, etc. But I didn't want to cloud it up, that way, just yet.

In your example case, the value of \$I_\text{B}=\frac{V_\text{BB}-V_\text{BE}} {R_\text{B}}\$ is set by the surrounding circuit. Assuming active mode for a moment, you'd expect \$I_\text{C}=\beta\cdot \frac{V_\text{BB}-V_\text{BE}} {R_\text{B}}\$, independent of the value of \$R_\text{C}\$. However, this implies that there is a voltage drop across \$R_\text{C}\$ of \$V_{\text{R}_\text{C}}=I_\text{C}\cdot R_\text{C}=\beta\cdot \frac{V_\text{BB}-V_\text{BE}} {R_\text{B}}\cdot R_\text{C}\$ and therefore that the BJT's collector voltage is \$V_\text{C}=V_\text{CC}-\beta\cdot \frac{V_\text{BB}-V_\text{BE}} {R_\text{B}}\cdot R_\text{C}\$. But if that calculation yields a value such that \$V_\text{C}\lt V_\text{BE}\$ here, then the BJT is somewhere between light and deep saturation and is no longer in active mode.

If you do the algebra, keeping things straight, this means that the BJT will be in saturation when, in this case, \$R_\text{C}\ge \frac{R_\text{B}}{\beta}\cdot\frac{V_\text{CC}-V_\text{BE}}{V_\text{BB}-V_\text{BE}}\$. From this, you can also tell what happens when you reduce the value of \$R_\text{C}\$. Assuming the BJT starts somewhere in saturation, then the transistor will move gradually towards active mode operation; moving into active mode when \$R_\text{C}\$ declines enough so that \$R_\text{C}\lt \frac{R_\text{B}}{\beta}\cdot\frac{V_\text{CC}-V_\text{BE}}{V_\text{BB}-V_\text{BE}}\$.