Binding Energy and Instability I believe that I am correct in saying that: (1) a nucleus constitutes a preferred configuration for nucleons as compared to a disassembled array of nucleons; and (2) nuclei decay so as to achieve more preferred nuclear configurations.  In saying this, I am arguing that an unstable nucleus, pre-decay, constitutes a preferred configuration for nucleons as compared to a disassembled array of nucleons.  Is this correct?
 A: Your statements are true. Nuclei prefer their stable configurations(lower energy) than their completely disassembled nucleons(having higher energy).
See this picture

Even though after Iron, adding more nucleons actually make the nuclei more unstable but still the stability is still more than for individual nucleons.
So yes, nuclei are always more stable than their separated individual nucleons.
A: Having a higher binding energy for constituent particles makes it less likely that any particle will escape.    These are the stable isotopes. "Preference" is a bit vague.
So point 2 is not exactly correct. An element can decay to one with shorter half life (a less preferred one?) in some circumstances. It may not stay like that for very long.
The unstable isotopes are a way off the red line on that chart in Houssains answer. All isotopes on the red line are relatively stable.
A: The problem is in words preferred configuration. I will try to argue -that there is nothing like preferred configuration.
Nucleons find themselves in some state at certain moment. The rule #1:


*

*they dont remember their history


which has impact on how the processes proceed - I mean specifically a decay law. There is - in every single moment $\Delta t$ - always the same probability to decay to something else, until it happens.
The rule #2:


*they dont see the future neither


which means, they do not decide for a preferred configurations. They are "only" and "all the time" confronted with all the possible outcomes (final states) - and 


*

*if the output is energetically possible (+ other conservation laws)

*if there is an interaction that can provide the change (strong, elmg, weak)
then a change can happen (remember rule #1, they dont remember how many times they have already rejected to go). The probability depends on the interaction strength, on the similarity of the initial and final states (wavefunctions).
To have a complete set, there must be a Rule 3:


*the interactions are time invariant
(not really 100%, it is CPT), which means, that the probability to convert from the initial to the final state is (when various factors are extracted) the same as from the final state to the initial one. 
Sounds strange. But the difference is that the photon from gamma decay or $e^- + \bar{\nu}$ from beta decay leave the decay place and go to $\infty$. They are not around anymore to tempt the nucleus to go back to the initial state. This has something to do with entropy and arrow of time - which would be (and potentially already is) a nice question here....
