Exact arithmetic complexity of Ryser's formula for computing permanent
Ryser: $n(n-2)2^{n-1} + n$ additions and $(n-1)(2^{n}-1)$ multiplications.

Ryser+Gray code: $n(2^{n}-2)$ additions and $(n-1)(2^{n}-1)$ multiplications.

Number of multiplications: For each nonempty subset of $[n]$, $n-1$ multiplications are used to multiply $n$ sums together.

Number of additions for Ryser: for each nonempty $S \subseteq [n]$, and for each $i \in [n]$ you compute $\sum_{j \in S} a_{ij}$ which uses $|S|-1$ additions. $\sum_{\emptyset \neq S \subseteq [n]}(|S|-1) = \sum_{k=1}^{n} (k-1) \binom{n}{k} = \sum_{k=1}^{n} k \binom{n}{k} - \sum_{k=1}^{n} \binom{n}{k} = n2^{n-1} - (2^{n}-1)$. 

Additions for Ryser+Gray: The Gray code version does not give you a smaller formula, but only a smaller circuit (which is still good, I just thought it was worth pointing out). Its savings require the re-use of previously computed quantities. For each $i \in [n]$, it does a Gray code over the nonempty sets $S \subseteq [n]$. Since there are $2^{n}-1$ such sets, and each transition of the Gray code involves a single addition/substraction, that gives the $2^{n}-2$.