Let $R_{\min }$ and $R_{\max }$ are the minimum and maximum resistances.
When all resistors are connected in parallel, the resultant resistance $R_p$ is
$$
\frac{1}{\mathrm{R}_{\mathrm{P}}}=\frac{1}{\mathrm{R}_1}+\ldots+\frac{1}{\mathrm{R}_{\mathrm{n}}}
$$
On multiplying both sides by $R_{\min }$ we get,
$$
\frac{\mathrm{R}_{\min }}{\mathrm{R}_{\mathrm{p}}}=\frac{\mathrm{R}_{\min }}{\mathrm{R}_1}+\frac{\mathrm{R}_{\min }}{\mathrm{R}_2}+\ldots+\frac{\mathrm{R}_{\min }}{\mathrm{R}_{\mathrm{n}}}
$$
Here, in RHS, there exist one term $\frac{R_{\text {min }}}{R_{\text {min }}}=1$ and other terms are positive, so we have
$$
\frac{\mathrm{R}_{\min }}{\mathrm{R}_p}=\frac{\mathrm{R}_{\min }}{\mathrm{R}_1}+\frac{\mathrm{R}_{\min }}{\mathrm{R}_2}+\ldots+\frac{\mathrm{R}_{\min }}{\mathrm{R}_{\mathrm{n}}}>1
$$
i.e., the resultant resistance $\left(R_p < R_{\text {min }}\right)$.
So, in parallel combination, the equivalent resistance of resistors is less than the minimum resistance available in combination of resistors.
Now, in series combination, the equivalent resistant is
$$
\mathrm{R}_{\mathrm{s}}=\mathrm{R}_1+\ldots+\mathrm{R}_{\mathrm{n}}
$$
Here, in RHS, there exist one term having resistance $\mathrm{R}_{\max }$
So, we get
$$
\begin{aligned}
\mathrm{R}_{\mathrm{s}} &=\mathrm{R}_1+\ldots+\mathrm{R}_{\max } \ldots+\ldots+\mathrm{R}_{\mathrm{n}} \\
\mathrm{R}_{\mathrm{s}} &=\mathrm{R}_1+\ldots \mathrm{R}_{\max }+\mathrm{R}_{\mathrm{n}} \\
&=\mathrm{R}_{\max }+\ldots\left(\mathrm{R}_1+\ldots+\mathrm{R}_{\mathrm{n}}\right.\\
\mathrm{R}_{\mathrm{s}} & \geq \mathrm{R}_{\max } \\
\mathrm{R}_{\mathrm{s}} &=\mathrm{R}_{\max }\left(\mathrm{R}_1+\ldots \mathrm{R}_{\mathrm{n}}\right)
\end{aligned}
$$
So, in series combination, the equivalent resistance is always greater than the maximum resistance $\left(\mathrm{R}_{\max }\right)$ available in combination of resistors. Physical interpretation


In Fig. (b), $R_{\text {min }}$ provides an equivalent route as in Fig (a) for current. But in addition there are $(\mathrm{n}-1)$ routes by the remaining $(\mathrm{n}-1)$ resistors. Current in Fig (b) is greater than current in Fig. (a). Effective resistance in Fig. (b) $ < \mathrm{R}_{\text {min. }}$. Second circuit evidently affords a greater resistance.


In the above figure, $R_{\max }$ provides an equivalent route as in Fig. (c) for current. Current in Fig.(d) < current in Fig. (c). Effective resistance in Fig. (d) $>\mathrm{R}_{\max }$. Second circuit evidently affords a greater resistance.