By $c$ character I mean the numbers $0,\dots,c-1$ and the blank symbol $b$, and by $n$ states I mean $n$ non-accepting states, reject and accept.
We can assume every $n$-state Turing machine has $(c+1)n$ transitions going to either another non-accepting state, reject, or accept, that is that it has a transition for every state and character on the tape. It can either move to the left, right or not shift each transition. I tried some myself to figure out how many Turing machines there were and got the two following incompatible results.
We have $(c+1)n$ transitions and each transition has $3(n+2)(c+1)$ different options. Encoding it as a 5 tuple I get $3n(n+2)(c+1)^2$, but taking from options for transitions for each transition gives me $((c+1)n)^{3(n+2)(c+1)}$
Is either of these correct and if so why?