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The formula for two resistors in parallel is:

$$\frac{1}{R_{p}} = \frac{1}{R_2} + \frac{1}{R_3}$$

Substitute the given values $R_2 = 20 \, \Omega$ and $R_3 = 30 \, \Omega$:

$$\frac{1}{R_{p}} = \frac{1}{20 \, \Omega} + \frac{1}{30 \, \Omega}$$

Find a common denominator, which is 60:

$$\frac{1}{R_{p}} = \frac{3}{60 \, \Omega} + \frac{2}{60 \, \Omega}$$ $$\frac{1}{R_{p}} = \frac{3+2}{60 \, \Omega} = \frac{5}{60 \, \Omega}$$

Now, invert the fraction to find $R_p$:

$$R_{p} = \frac{60 \, \Omega}{5} = 12 \, \Omega$$

The parallel combination $R_p$ is in series with $R_1$.
The formula for resistors in series is:

$$R_{total} = R_1 + R_{p}$$

Substitute the values $R_1 = 10 \, \Omega$ and $R_p = 12 \, \Omega$:

$$R_{total} = 10 \, \Omega + 12 \, \Omega$$ $$R_{total} = 22 \, \Omega$$

The total resistance of the circuit is $\text{22 } \Omega$.