You are given thirty two capacitors each having capacity 4 $\mathrm{\mu }$F. How do you connect all of them to prepare a composite capacity having capacitance 8 $\mathrm{\mu }$F?

Let m condensers are joined in series and n such groups are in parallel.
Capacity of m condensers connected in series
$\frac{1}{{\mathrm{C}}^{\mathrm{\prime }}}=\frac{1}{{\mathrm{C}}_1}+\frac{1}{{\mathrm{C}}_2}+\dots =\frac{\mathrm{m}}{\mathrm{C}}\text{ or }{\mathrm{C}}^{\mathrm{\prime }}=\frac{\mathrm{C}}{\mathrm{m}}=\frac{4}{\mathrm{m}}$
When n such groups are connected in parallel, then
${\mathrm{C}}_{\mathrm{eff}}={\mathrm{nC}}^{\mathrm{\prime }}=\frac{4\mathrm{n}}{\mathrm{m}}$
Given that $C_{\mathrm{eff}}=8,\therefore 8=\frac{4\mathrm{n}}{\mathrm{m}}$
Or     $\frac{\mathrm{n}}{\mathrm{m}}=2\text{ or }\mathrm{n}=2\mathrm{m} ...\left(1\right)$
Further                 $\mathrm{mn}=32 ....\left(2\right)$
Now    $\mathrm{m}\times 2\mathrm{m}=32\text{ or }\mathrm{m}=4$
and     n = 2 x 4 = 8