The principle of inclusion and exclusion: How many permutations of the set $\{1, 2,. . . , 8\}$ do not leave any even number in its place? I am supposed to solve the following problem:
How many permutations of the set $\{1, 2,. . . , 8\}$ do not leave any even number in its place?
What I tried:
$$8!-\left ( \binom{8}{1}7!-\binom{8}{ 2}6!+\binom{8}{3}5!-\binom{8}{4}4! \right )$$
But I know that this is incorrect. 
Can anyone tell me why?

Let $S_j$ be the set of arrangements with $2j$ left in place. Then
$$
N_k=\overbrace{\ \ \ \binom{n}{k}\ \ \ }^{\substack{\text{number of}\\\text{ways to pick}\\\text{the $k$ fixed}\\\text{numbers}}}\overbrace{\vphantom{\binom{n}{k}}(2n-k)!}^{\substack{\text{arrangements}\\\text{of the}\\\text{remaining}\\\text{numbers}}}
$$
According to the Generalized Inclusion-Exclusion Principle, the number of arrangements in none of the $S_j$ is
$$
\sum_{k=0}^n(-1)^kN_k
$$
for $n=4$, this gives
$$
\binom{4}{0}8!-\binom{4}{1}7!+\binom{4}{2}6!-\binom{4}{3}5!+\binom{4}{4}4!=24024
$$
