Question: Does there exist nontrivial homomorphism between symmetric group $S\_n$ to set of integer $\mathbb Z$?

If yes, how many?

If not, then why?

Answer: 
Noticing that there are very few elements of finite order in $\mathbb Z$ helps. In particular, since a homomorphism sends elements of finite order to elements of finite order, and in a finite group $G$ all elements are of finite order, there is really very little freedom in what a homomorphism $G\to \mathbb Z$ can do.
If $\phi$ is a group homomorphism between two groups $G$ to $H$ , then for any $a \in G$ , $o(\phi(a)) | o(a) \ldots (\*)$ . Since we are considering homomorphisms from $S\_n$ to $\Bbb Z$ and $S\_n$ is a finite group, any element in $S\_n$ is a divisor of $n!$ and hence finite . Thus its image element must also be of finite order (due to $(\*)$), but in $\Bbb Z$ , 0 is the only element of finite order, thus all the elements in $S\_n$ should be mapped to 0 i.e. the trivial 0-homomorphism is the only homomorphism from $S\_n$ to $\Bbb Z$ .

Note that here we are not using any special property of $S\_n$ , in fact the statement is true for any finite group .