If two or more capacitors $C_1$, $C_2$, ... $C_n$ are connected in series they are equivalent with a capacitor with $$C_{eff}=\frac{1}{\frac{1}{C_1}+\frac{1}{C_2}+...+\frac{1}{C_n}}$$ That means we can replace one of the $n$ capacitors with an equivalent capacitor with capacitance $C_{eff}$ and replace the other capacitors with short-circuits (wires). When we have only two capacitors connected in parallel, we can replace them with a single capacitors with capacitance $$C_{eff}=\frac{C_1 C_2}{C_1+C_2}$$

For instance, considering the circuit in Fig. 1, capacitors $C_1$ and $C_2$ are connected in series. Therefore, we can keep one the capacitors, say $C_1$, replace its value with $C_{eff}=\frac{C_1 C_2}{C_1+C_2}$, 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 with a capacitor with $$C_{eff}=C_1+C_2+...+C_n$$ That means we can replace one of the $n$ capacitors with an equivalent capacitor with capacitance $C_{eff}$ and remove all the other capacitors.

For instance, considering the circuit in Fig. 2, capacitors $C_3$, $C_4$, and $C_5$ are connected in parallel. Therefore, we can keep one the capacitors, say $C_3$, replace its value with $C_{eff}=C_3+C_4+C_5$, 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 $C_4$.