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

We are asked to arrange the four given resistors 2 $$\Omega$$, 4 $$\Omega$$, 6 $$\Omega$$ and 8 $$\Omega$$ so that the overall or equivalent resistance becomes $$\dfrac{46}{3}\,\Omega$$. Let us test Option C, because it places 2 $$\Omega$$ and 4 $$\Omega$$ in parallel, while 6 $$\Omega$$ and 8 $$\Omega$$ are first placed in series.

First, we form the parallel combination of 2 $$\Omega$$ and 4 $$\Omega$$. For two resistors in parallel we use the formula

$$\frac{1}{R_{\text{parallel}}}= \frac{1}{R_1}+\frac{1}{R_2}.$$

Substituting $$R_1 = 2\,\Omega$$ and $$R_2 = 4\,\Omega$$, we get

$$\frac{1}{R_{24}} = \frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4}.$$

Now we invert to find the resistance value itself:

$$R_{24} = \frac{4}{3}\;\Omega.$$

Next, we form the series combination of 6 $$\Omega$$ and 8 $$\Omega$$. For resistors in series we simply add them:

$$R_{\text{series}} = R_1 + R_2.$$

So, with $$R_1 = 6\,\Omega$$ and $$R_2 = 8\,\Omega$$, we have

$$R_{68} = 6 + 8 = 14\;\Omega.$$

We now have two equivalent resistances: $$R_{24}= \dfrac{4}{3}\,\Omega$$ and $$R_{68}=14\,\Omega$$. According to Option C these two combinations are connected in series, therefore their resistances add:

$$R_{\text{eq}} = R_{24} + R_{68} = \frac{4}{3} + 14 = \frac{4}{3} + \frac{42}{3} = \frac{46}{3}\;\Omega.$$

This final value matches exactly the required $$\dfrac{46}{3}\,\Omega$$. Hence, the correct answer is Option C.