Calculate the equivalent resistance between the points $A$ and $B$ in the following combination.

$(11 \Omega)$ The given circuit can be redrawn by identifying the nodes connected by wires. Points $A$ and $C$ are at the same potential, and points $E$ and $F$ are connected to $D$ via a wire, effectively making them parallel.
It is clear that the resistances $20 \Omega$ (between $C$ and $D$), $10 \Omega$ (between $D$ and $E$), and $20 \Omega$ (between $E$ and $F$) are in a parallel combination between the effective nodes.
If $R_{1}$ is the effective resistance of these three parallel resistors, then:
$\frac{1}{R_{1}} = \frac{1}{20} + \frac{1}{10} + \frac{1}{20}$
$\frac{1}{R_{1}} = \frac{1 + 2 + 1}{20} = \frac{4}{20} = \frac{1}{5}$
Therefore, $R_{1} = 5 \Omega$.
Now, the resistor $R_{1}$ and the $6 \Omega$ resistor are in series. Therefore, the equivalent resistance $R$ between $A$ and $B$ is:
$R = R_{1} + 6 \Omega$
$R = 5 \Omega + 6 \Omega = 11 \Omega$.