I am trying to calculate the time complexity of the $k$-opt algorithm. I started off by first calculating the number of ways of reconnecting the $k$-cycle given that $k$ non-consecutive arcs are removed. The answer is $(k-1)!\,2^{k-1}$.
Next I need to know the number of ways $k$ non-consecutive arcs can be removed in a $k$-cycle. The cycle must be of length $N \ge 2k$. How many ways are there of picking $k$ arcs such that no two arcs are adjacent?
When you use terminology from a specialized field (in this case the study of the travelling salesperson problem) in a question, it's usually a good idea to define your terms, at least those whose definition can't immediately be found by searching, and especially those for which, as in this case, a search leads to a different definition than the one you're using.
I take it you're referring to the method of optimizing a TSP solution by removing $k$ non-consecutive arcs and reconnecting the remaining parts of the tour in all possible ways. If so, wherever you have $k$-cycle it should be $N$-cycle; clearly it's impossible to remove $k$ non-consecutive arcs in a $k$-cycle.
Your result $(k-1)!2^{k-1}$ for the first part of the calculation is correct. To find the number of ways of picking $k$ non-consecutive arcs in an $N$-cycle, pick pairs of adjacent arcs instead and then from each pair use the arc that's, say, clockwise from the other. First pick one pair to get from a cycle to a linear chain; then pick $k-1$ pairs in the chain; then correct for the fact that any of the $k$ pairs could have been picked first. That gives $N$ options for the position of the first pair, $\binom{N-k-1}{k-1}$ options for the positions of the remaining pairs, and $k$ options for which pair to pick first, so the result is
$$\frac Nk\binom{N-k-1}{k-1}=\frac{N(N-k-1)!}{k!(N-2k)!}\;.$$
