Given a PRAM may use arbitrarily many processors, why is Hamiltonian Cycle not in NC? In my parallel algorithms class, the PRAM model is described as having an "arbitrary number of processors, bounded by some polynomial in the input size."

I think that this may be missing a constraint.  For example, this would imply that finding a Hamiltonian cycle is in NC:

If I give all of the permutations of the vertices of a graph as input,
then I can assign a processor to each permutation, and each processor can 
then check for the appropriate edges in O(|E|) time, which would mean that 
finding a Hamiltonian cycle would be in NC.

So is there some constraint that is missing from the definition we are using in class, or am I missing something else?

Edit

For clarification:  Consider a classical TM, which accepts as input a sane representation of a graph, followed by all of the permutations of a graph's vertices.  It then performs the best currently known algorithm for finding Hamiltonian Cycles, and as output writes out a permutation of the vertices describing such a cycle.

The algorithm would perform in time linear in the input size, but in time factorial in the order of the input graph.  (The extra inputs don't seem to count as computations for a classic TM.)

In the PRAM model (as I have described it), the algorithm would perform in time linear in both the input size, and in the order of the graph.  (The extra inputs seem to count as computations for the PRAM model...  Which makes me think that "computation" means something subtly different here...)

If this is the way it is supposed to work: awesome.  If not, than am I missing some constraints on what the allowed inputs are?
If I give all of the permutations of the vertices of a graph as input


That's not the input for Hamiltonian Cycle; you only get the graph.


  then I can assign a processor to each permutation


Since there are $n! \in \omega(c^k)$ many permutations, you don't have that many processors.

Here's the twist: if you were allowed to use all permutations as input, you'd of course have polynomially many processors in $n!$ at your disposal. But since your input is the graph you only get polynomially many in $n$.