<s>[INST] Find the remainder when $\frac{13!}{7!}$ is divided by $17$. How to find the remainder when $\frac{13!}{7!}$ is divided by $17$?
I started with Wilson's Theorem which states that every prime $p$ divides $(p-1)!+1$. That is $(p-1)!=-1 mod(p)$. 
$16!=-1 mod(17)$
Kindly help me how to get to $\frac{13!}{7!}$.
 [/INST] By Wilson's theorem $16!\equiv -1\pmod{17}$ and since $17$ is a prime of the form $8k+1$ and the square roots of $-1$ in $\mathbb{F}_{17}^*$ are $4$ and $13$, $8!\equiv 13\pmod{17}$ and $7!\equiv 8\pmod{17}$. By using the notation $\frac{1}{a}$ for the inverse of $a$,
$$ \frac{13!}{7!}\equiv\frac{16!}{(-1)(-2)(-3)8}\equiv\frac{1}{48}\equiv\frac{1}{-3}\equiv -6\equiv\color{red}{11}\pmod{17}.$$
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