Q:

Is the set of non-deterministic Turing machines countable?

We know that deterministic TMs are countable (enumeration). Does the same hold for NTMs? Are TMs and NTMS equinumerous?

A:

Why is the class of DTM countable? Because you can code every DTM as a string (the program) over some alphabet. Now you can sort these strings lexicographicaly. Then the function 
$$f(i\in \mathbb{N}):=\text{the $i$th string in the sorted sequence of DTMs},$$
is a bijection between the natural numbers and the set of DTMs. Hence the set of DTMs is countable.
Now here is the point. You can argue the same for NTMs. Hence the set of DTMs and NTMs are both countable and therefore equinumerous.

