Constructing a Simple Group with a Specific Number of Sylow $p$-subgroups I've been studying for my final exams, and I came across the following question:

If possible, give an example of a simple group $G$ with $n>1$ Sylow $p$-subgroups such that the order of $G$ does not divide $n!$. If not possible, briefly explain why.

Now, it didn't take me long to think of the following "numerical" example: a group of order $60=2^2\cdot3\cdot5$ where there are $3$ Sylow $2$-subgroups. My question is whether or not such a group exists, and if it does, is there any simple way to describe these groups based on knowing how the Sylow structure is built?
In trying to describe the above group, I think I showed that it can't exist. This is because there either have to be $4$ Sylow 3-subgroups or $10$ Sylow 3-subgroups. The second case leads to a contradiction since then there would have to be $k$ Sylow 5-subgroups such that $k(5-1)=30$, and likewise, in the first case, we find we'd have to have 13 Sylow 5-subgroups, also a contradiction. Is all of this correct and is there an easy way to determine whether such a group exists or not?
 A: The answer is no.
Let $C$ be the set of sylow $p$ subgroups of $G$. It is easy to verify that the map $G\times C\rightarrow C$ that sends $(x,H)$ to $xHx^{-1}$ is an action of $G$ on $C$. By the orbit stabilizer theorem, we know that:
$$\forall H\in C[|C||N(H)|=|G|]$$
Since $1<|C|$, we deduce that $|N(H)|<|G|$ for all $H\in C$....(1) 
By sylow's theorem we know that the function $T_x:C\rightarrow C$ that sends $H$ to $xHx^{-1}$ is a permutation of $C$. Now consider the homomorphism $\phi:G\rightarrow S_C$ that sends $x$ to $T_x$. 
Claim:  $\phi$ is an injection.
Proof: We know that $\operatorname{Im}\phi\cong G/\ker\phi$. Thus, it suffices to show that $|\ker\phi|=1$. This is easy to see because $\ker\phi\unlhd G$, since $G$ is simple, therefore either $\ker\phi=\{e\}$ or $\ker\phi=G$. Let $H\in C$, it is easy to verify that $\ker\phi \leq N(H)$. Thus, if $\ker\phi=G$, we would conclude that $N(H)=G$. However, we know from (1) that $N(H)$ is a proper subgroup of $G$. Therefore, we finally deduce that $\ker\phi=\{e\}$. 
Thus, $G$ is isomorphic to a subgroup of $S_C$ (The group of permutations of $C$). Hence, $|G|\,\Big|\,|S_C|!=n!$
