How many Turing machines are there with $c$ characters and $n$ states?
The answer is $(3(c+1)(n+2))^{(c+1)n}$. You get this by considering all functions of the type $(Q \setminus F) \times \Gamma \to Q \times \Gamma \times \{L, R, N\}$. Where $Q$ is the set of states, $F$ is the set of final states, $\Gamma$ is the alphabet including the blank character, and $L$, $R$, $N$ are the left shift, right shift, and no shift respectively. You can view this as a problem of picking $|(Q \setminus F) \times \Gamma| = (c+1)n$ elements from a set of cardinality $|Q \times \Gamma \times \{L, R, N\}| = 3(c+1)(n+2)$ with replacement.