Three capacitors each of capacity 4 $\mu F$ are to be connected in such a way that the effective capacitance is 6 $\mu F$. This can be done by -

Check each arrangement (C = 4 $\mu F$):
All in series: $\dfrac{C}{3} = \dfrac{4}{3}\ \mu F$
All in parallel: $3C = 12\ \mu F$
Two in series and this pair in parallel with the third: $\dfrac{C}{2} + C = 2 + 4 = 6\ \mu F$

Two in parallel and this pair in series with the third: $\dfrac{8 \times 4}{8 + 4} = \dfrac{8}{3}\ \mu F$
So $C_{AB} = (2 + 4)\ \mu F = 6\ \mu F$ is obtained by connecting two in series and one in parallel.