In the circuit shown in the figure, the input voltage $V_i$ is 20 V, $V_{BE} = 0$ and $V_{CE} = 0$. The values of $I_B$, $I_C$ and $\beta$ are given by

Given $V_{CE} = 0$, $I_C = \frac{V_{CC} - V_{CE}}{R_C} = \frac{20 - 0}{4 \times 10^3} = 5 mA$. Given $V_{BE} = 0$, $V_i = V_{BE} + I_B R_B \implies 20 = 0 + I_B(500 \times 10^3) \implies I_B = \frac{20}{500 \times 10^3} = 40 \mu A$. Then $\beta = \frac{I_C}{I_B} = \frac{5 \times 10^{-3}}{40 \times 10^{-6}} = 125$.