Physics Capacitance helpWatch
How does changing the number of Capacitors affect charge and energy storage as well as potential difference in series and parallel layouts?
For capacitors in parallel, the sum would be equal to all the capacitance added up. A 2F and 3F in parallel sums to 5F, and hence the circuit can store approximately 5 units charge for each unit potential difference EMF (assuming none of that internal resistance hoobleyjoobley).
In series, the p.d. is split up between the capacitors, and hence each of them stores less charge than they would in the above situation. Taking that in to account, the total capacitance is equal to (1/2 + 1/3)^-1, so, (1/C1 + 1/C2)^-1, and you can do this for all the capacitors you may desire, if that's your kink.
Either ways, I hope I helped. Joinedup had a far better solution and the website provides much more insight.