Which of the following combination of 7 identical capacitors each of \(2 \mu \mathrm{F}\) gives a capacitance of \(\frac{10}{11} \mu \mathrm{F}\) ?

For \(\mathrm{n}\) identical capacitors connected in series, the equivalent capacitance is, \(\mathrm{C}_{\mathrm{s}}=\frac{\mathrm{C}}{\mathrm{n}}\)
Similarly, for \(\mathrm{m}\) identical capacitors connected parallel to each other, the equivalent capacitance is, \(\mathrm{C}_{\mathrm{p}}=\mathrm{mC}\)
Assuming the two combinations are connected in series, the net capacitance,
\(\frac{1}{\mathrm{C}_{\text {net }}} =\frac{1}{\mathrm{mC}}+\frac{\mathrm{n}}{\mathrm{C}}=\frac{11}{10} \mu \mathrm{F} \ldots(\because \mathrm{C}_{\text {net }}\) \(=\frac{10}{11} \mu \mathrm{F}) \)
\( \therefore \text { for } \mathrm{C} =2 \mu \mathrm{F}, \)
\( \frac{11 \times 2}{10} =\frac{1}{\mathrm{~m}}+\mathrm{n}\)
\(\therefore \frac{1}{\mathrm{~m}}+\mathrm{n}=\frac{11}{5}...(i)\)
Substituting the values for \(\mathrm{m}\) and \(\mathrm{n}\) in equation (i) from each option, the correct answer can be found to be (A).