Let's start by understanding the process of dividing the wire and recombining its parts to form the final configuration. Initially, we have a wire with a resistance of $$20 \Omega$$. This wire is divided into 10 equal parts, each part then has a resistance of:

$$\frac{20 \Omega}{10} = 2 \Omega$$

Since each part has the same length and presumably the same material and cross-sectional area, then each part will have the same resistance of $$2 \Omega$$.

When two parts are connected in parallel, the equivalent resistance, $$R_{\text{parallel}}$$, of this configuration can be calculated using the formula for two resistors in parallel:

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

Given that $$R_1 = R_2 = 2 \Omega$$ (since the parts are identical), we have:

$$\frac{1}{R_{\text{parallel}}} = \frac{1}{2 \Omega} + \frac{1}{2 \Omega} = \frac{2}{2 \Omega}$$

This simplifies to:

$$\frac{1}{R_{\text{parallel}}} = \frac{2}{2 \Omega} = \frac{1}{\Omega}$$

From which it follows that:

$$R_{\text{parallel}} = 1 \Omega$$

Now, since the original wire was divided into 10 equal parts, and pairs of these parts are connected in parallel, this results in $$\frac{10}{2} = 5$$ pairs. Each of these pairs has an equivalent resistance of $$1 \Omega$$.

Finally, these pairs are all connected in series. The total resistance of resistors in series is simply the sum of their individual resistances. Therefore, the equivalent resistance of the final configuration, $$R_{\text{series}}$$, is:

$$R_{\text{series}} = 5 \times R_{\text{parallel}} = 5 \times 1 \Omega = 5 \Omega$$

So, the equivalent resistance of the final combination is $$5 \Omega$$.