If two or more capacitors — $C_1$, $C_2$, ..., $C_n$ — are connected in series, they are equivalent to a single capacitor with $$\begin{equation}C_{eff}=\frac{1}{\frac{1}{C_1}+\frac{1}{C_2}+...+\frac{1}{C_n}}\end{equation}$$ That means we can replace one of the $n$ capacitors with an equivalent capacitor of capacitance $C_{eff}$ and short-circuit (wire) the remaining capacitors.

For instance, consider the circuit in Fig. 1, in which capacitors $C_1$ and $C_2$ are connected in series. In this circuit, we can keep one capacitor, say $C_1$, set its value to $C_{eff}=\frac{1}{\frac{1}{C_1}+\frac{1}{C_2}}$, and replace capacitor $C_2$ with a wire.

If two or more capacitors — $C_1$, $C_2$, ..., $C_n$ — are connected in parallel, they are equivalent to a single capacitor with $$\begin{equation}C_{eff}=C_1+C_2+...+C_n\end{equation}$$ That means we can replace one of the $n$ capacitors with an equivalent capacitor of capacitance $C_{eff}$ and remove the other capacitors.

For instance, consider the circuit in Fig. 2, in which capacitors $C_3$, $C_4$, and $C_5$ are connected in parallel. In this circuit, we can keep one of the capacitors, say $C_3$, set its value to $C_{eff}=C_3+C_4+C_5$, and remove capacitors $C_4$ and $C_5$ from the circuit. Similarly, we could keep $C_4$ and remove $C_3$ and $C_5$, or keep $C_5$ and remove $C_3$ and $C_4$.