This was a question from 'Brilliant.org' and I could not find a solution. The question is
Find the number of 64 tuples $(x_0,x_1,...x_{63})$ such that $x_0,x_1,x_2,...x_{63}$ are distinct elements of $\mathbb{Z}_{2017}$ and $$x_0+x_1+2x_2+3x_3+...+63x_{63}=0\,\,\, \text{mod} \,\, 2017.$$ If your answer is of the form $$n!\left(\binom{m}{n} - m\right),$$ submit $m-n.$
I was trying to write a generating function. But I don't know how to impose the distinct elements condition. Any approach is appreciated!
We will first prove the following lemma.
Lemma. Fix a prime $p$ and define the function $f(k)$ on positive integers by the conditions $$\begin{aligned} & f(1, p)=0 \\ & f(k, p)=\frac{(p-1) !}{(p-k) !}-k f(k-1, p) \quad(k>1) . \end{aligned}$$
Then for any positive integers $a_1, \ldots, a_k$ with $a_1+\cdots+a_k<p$, there are exactly $f(p)$ solutions to the equation $a_1 x_1+\cdots+$ $a_k x_k=0$ with $x_1, \ldots, x_k \in \mathbb{F}_p$ nonzero and pairwise distinct. Proof. We check the claim by induction, with the base case $k=1$ being obvious. For the induction step, assume the claim for $k-1$. Let $S$ be the set of $k$-tuples of distinct elements of $\mathbb{F}_p$; it consists of $\frac{p !}{(p-k) !}$ elements. This set is stable under the action of $i \in \mathbb{F}_p$ by translation: $$\left(x_1, \ldots, x_k\right) \mapsto\left(x_1+i, \ldots, x_k+i\right)$$ Since $0<a_1 \cdots+a_k<p$, exactly one element of each orbit gives a solution of $a_1 x_1+\cdots+a_k x_k=0$. Each of these solutions contributes to $f(k)$ except for those in which $x_i=0$ for some $i$. Since then $x_j \neq 0$ for all $j \neq i$, we may apply the induction hypothesis to see that there are $f(k-1, p)$ solutions that arise this way for a given $i$ (and these do not overlap). This proves the claim.
To compute $f(k, p)$ explicitly, it is convenient to work with the auxiliary function $$g(k, p)=\frac{p f(k, p)}{k !}$$ by the lemma, this satisfies $g(1, p)=0$ and $$\begin{aligned} g(k, p) & =\left(\begin{array}{c} p \\ k \end{array}\right)-g(k-1, p) \\ & =\left(\begin{array}{c} p-1 \\ k \end{array}\right)+\left(\begin{array}{c} p-1 \\ k-1 \end{array}\right)-g(k-1, p) \quad(k>1) . \end{aligned} $$ By induction on $k$, we deduce that $$ \begin{aligned} g(k, p)-\left(\begin{array}{c} p-1 \\ k \end{array}\right) & =(-1)^{k-1}\left(g(1, p)-\left(\begin{array}{c} p-1 \\ 1 \end{array}\right)\right) \\ & =(-1)^k(p-1) \end{aligned} $$ and hence $g(k, p)=\left(\begin{array}{c}p-1 \\ k\end{array}\right)+(-1)^k(p-1)$. We now set $p=2017$ and count the tuples in question. Define $c_0, \ldots, c_{63}$ as in the first solution. Since $c_0+\cdots+c_{63}=p$, the translation action of $\mathbb{F}_p$ preserves the set of tuples; we may thus assume without loss of generality that $x_0=0$ and multiply the count by $p$ at the end. That is, the desired answer is $$\begin{aligned} 2017 f(63,2017) & =63 ! g(63,2017) \\ & =63 !\left(\left(\begin{array}{c} 2016 \\ 63 \end{array}\right)-2016\right). \end{aligned}$$
This solution is based on a variation found here.