The Number of quotient group of $S_4$ (the symmetric group of 4 symbols) up to isomorphism The Number of quotient group of $S_4$ (the symmetric group of $4$ symbols) up to isomorphism is/are
a)$1$ b)$2$ c)$3$ d)$4$

A quotient group (up to isomorphism) is precisely the image of a homomorphism. So the question is asking how many groups $G$ there are such that there is a surjective homomorphism $S_4 \to G$; equivalently, we're looking for the number of non-isomorphic normal subgroups of $S_4$.
The normal subgroups of $S_4$ are $A_4$ (of course, being index $2$); $V_4$ by sheer luck; $\{e\}$; and $S_4$ itself. We can work this out by listing the conjugacy class sizes in $S_4$ (using the fact that cycle type = conjugacy class), and then using the fact that normal subgroups are precisely unions of conjugacy classes and must have size dividing $|S_4|$.

There's a lovely trick, by the way, to make $S_4$ act on three elements. Consider the usual action of $S_4$ on four elements $\{a,b,c,d\}$. This action tells us how to make a new action of the same group $S_4$ on three elements, by the cunning move of letting the three elements be $\mathcal{S} = \{ \{(a,b), (c,d)\}, \{(a,c),(b,d)\}, \{(a,d),(b,c)\}\}$ the collection of partitions of $\{a,b,c,d\}$ into two pairs.
That yields a homomorphism: send $\sigma \in S_4$ to the corresponding element of $\mathrm{Sym}(\mathcal{S})$. For example, the transposition $(a,b) \in S_4$ gets sent to $$[(a,b)] \mapsto [(a,b)]; \ [(a,c)] \mapsto [(a,d)]; \ [(a,d)] \mapsto [(a,c)]$$
where I have written $[(a,b)]$ as a shorthand for the set $\{(a,b), (c,d)\}$, and so on.
