Is there any formula to calculate the number of normal subgroups of $S_n$?
Suppose i have an answer to this question it is easy to answer how many homomorphism is there from $S_n$ to any other group. Because each normal subgroup defines a kernel for a homomorphism...
$S_n$ has exactly $2$ normal subgroups for $n = 2$, $4$ normal subgroups for $n = 4$, and $3$ normal subgroups for every other $n$ - these are $1$, $S_n$, and $A_n$.
For $n \le 4$ this can be done by hand (note that the Sylow $2$-subgroup of $A_4$ is normal in $A_4$, hence characteristic in $A_4$, hence normal in $S_4$), and for $n \ge 5$ this follows from simplicity of $A_n$ (if $H \unlhd S_n$, $H \not \supseteq A_n$, then $H \cap A_n \unlhd A_n \implies H \cap A_n = 1 \implies |H| = [H : H \cap A_n] = [HA_n : A_n] = [S_n : A_n] = 2$, but a subgroup of order $2$ is normal iff it is contained in the center, and $Z(S_n) = 1$).