Time Constant of a circuit - capacitors in parallel or series? 
When I charge the circuit as shown where the switch is at A, the p.d. across C1 is 10V, which is correct. 
When I discharge the circuit, however, by changing the switch S from A to B, the total capacitance of the circuit is 200μF as C1 and C2 are in parallel. 
This must mean the time constant of the discharging circuit is 
$$200 \times 10^{-6} \times 100 \times 10^3 = 20s$$
In the solution, however, the time constant of the discharging circuit is quoted as 5 seconds, where it states C1 and C2 are in series when the switch is at B. 
Who is right, the solutions or me?
 A: 
the total capacitance of the circuit is 200μF as C1 and C2 are in
  parallel.

They're not in parallel for either switch position.
When the switch is in position A, C2 and the 100k resistor are in series but one end of the resistor is 'dangling' so there is no path for current through the series combination.
When the switch is in position B, C1 is placed in series with the C2 + 100k series RC combination.  This should be obvious since there is only one path for current, all three circuit elements have identical current.
If the capacitors were parallel connected, C1 and C2 would 'split' the current through the 100k resistor but clearly, all of the current through the resistor is through either capacitor and so, the capacitors are series connected.
As you already know, the equivalent capacitance of 2 identical series connected capacitors is 1/2 the individual capacitance and thus
$$C_{eq} = 50\mu\mathrm{F}$$
and the time constant is
$$\tau = 100\mathrm{k\Omega}\cdot 50\mu\mathrm{F} = 5\mathrm{s}$$
