How can three resistors of resistances $2 \,\Omega$,$3 \,\Omega$,and $6 \,\Omega$ be connected to give a total resistance of $4 \,\Omega$?

(N/A) We have three resistors with resistances $R_1 = 2 \,\Omega$,$R_2 = 3 \,\Omega$,and $R_3 = 6 \,\Omega$.
To obtain a total resistance of $4 \,\Omega$,we connect the $3 \,\Omega$ and $6 \,\Omega$ resistors in parallel,and then connect this combination in series with the $2 \,\Omega$ resistor.
First,calculate the equivalent resistance $(R_p)$ of the $3 \,\Omega$ and $6 \,\Omega$ resistors connected in parallel:
$\frac{1}{R_p} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2}$
$R_p = 2 \,\Omega$
Next,connect this equivalent resistance $(R_p)$ in series with the $2 \,\Omega$ resistor:
$R_{total} = R_p + 2 \,\Omega = 2 \,\Omega + 2 \,\Omega = 4 \,\Omega$
Thus,the total resistance of the circuit is $4 \,\Omega$.