If you are provided a set of resistances 2Ω, 4Ω, 6Ω and 8Ω. Connect these resistances, so as to obtain an equivalent resistance of $\frac{46}{3}\mathrm{\Omega }$.

The given value of resistances are 2Ω, 4Ω, 6Ω and 8Ω.

The required value of combination is 46/3Ω.

In order to achieve the above mentioned values of resistance from given resistances, we will connect 2Ω and 4Ω resistance in parallel, then join 6 Ω and 8 Ω resistance in series with the combination.

The circuit diagram for connection is shown below.

$\therefore {\mathrm{R}}_{\mathrm{eq}}=(2\parallel 4)+6+8=\frac{2\times 4}{2+4}+14=\frac{46}{3}\mathrm{\Omega }$

Thus, resistance of 2Ω and 4Ω are in parallel with 6Ω and 8Ω in series combination.