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

We want an equivalent capacitance of $$6\,\mu\text{F}$$ using three identical capacitors, each having $$C = 4\,\mu\text{F}$$. To decide which connection gives this value, we recall two standard results for combining capacitors:

For capacitors connected in parallel, the capacitances simply add. The formula is

$$C_{\text{parallel}} = C_1 + C_2 + C_3 + \dots$$

For capacitors connected in series, the reciprocals add. The formula is

$$\frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \dots$$

We now test each option one by one.

1. All three in series. Using the series formula we have

$$\frac{1}{C_{\text{eq}}} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4},$$

so $$C_{\text{eq}} = \frac{4}{3}\,\mu\text{F} \approx 1.33\,\mu\text{F}.$$ This is far below $$6\,\mu\text{F},$$ so this arrangement does not work.

2. All three in parallel. Using the parallel formula we get

$$C_{\text{eq}} = 4 + 4 + 4 = 12\,\mu\text{F},$$

which is double the desired value. Hence this option is also unsuitable.

3. Two in parallel and the third in series with the combination. First, put two capacitors in parallel:

$$C_{\text{parallel}} = 4 + 4 = 8\,\mu\text{F}.$$

Now place this $$8\,\mu\text{F}$$ in series with the remaining $$4\,\mu\text{F}$$ capacitor. Using the series formula,

$$\frac{1}{C_{\text{eq}}} = \frac{1}{8} + \frac{1}{4} = \frac{1}{8} + \frac{2}{8} = \frac{3}{8},$$

so $$C_{\text{eq}} = \frac{8}{3}\,\mu\text{F} \approx 2.67\,\mu\text{F}.$$ This is still not the required $$6\,\mu\text{F}.$$

4. Two in series and the third in parallel with the combination. First, connect two capacitors in series. Using the series formula,

$$\frac{1}{C_{\text{series}}} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2},$$

which gives

$$C_{\text{series}} = 2\,\mu\text{F}.$$

Now place this $$2\,\mu\text{F}$$ result in parallel with the third $$4\,\mu\text{F}$$ capacitor. Using the parallel formula,

$$C_{\text{eq}} = 2 + 4 = 6\,\mu\text{F}.$$

This matches the required effective capacitance exactly.

Since only the arrangement with two capacitors in series and one in parallel yields $$6\,\mu\text{F},$$ that is the correct choice.

Hence, the correct answer is Option D.