Brazilian Mathematical Olympiad

Let $a = f(0, 0, 0, \dots, 0)$, $b = f(1, 0, 0, \dots, 0)$ and $c = f(2, 0, 0, \dots, 0)$, say $a = (a_1, a_2, \dots, a_{2000})$, $b = (b_1, b_2, \dots, b_{2000})$ and $c = (c_1, c_2, \dots, c_{2000})$. It's easy to conclude that there exists a unique $1 \le i_0 \le 2000$ such that $a, b$ and $c$ differ at position $i_0$, i.e., $a_{i_0} \ne b_{i_0} \ne c_{i_0} \ne a_{i_0}$. This also gives $i_0 \le 1000$. Another easy argument shows that if $x_0 = f(0, d_2, d_3, \dots, d_{2000}) = (x_1, x_2, \dots, x_{2000})$, then $x_{i_0} = a_{i_0}$ and also if $x_1 = f(1, e_2, e_3, \dots, e_{2000}) = (y_1, y_2, \dots, y_{2000})$, then $y_{i_0} = b_{i_0}$.
A similar argument shows the following: let
$$a_t = f(0, 0, \dots, 0, \dots, 0) = (a_1, a_2, \dots, a_{2000}),$$
$$b_t = f(0, 0, \dots, 1, \dots, 0) = (b_1, b_2, \dots, b_{2000}),$$
$$c_t = f(0, 0, \dots, 2, \dots, 0) = (c_1, c_2, \dots, c_{2000}),$$
where $t \le 1000$ is the position of 1 and 2, and $x = f(x_1, x_2, \dots, x_{2000}) = (y_1, y_2, \dots, y_{2000})$. Therefore $y_{i_t} = a_{i_t}$ if $x_t = 0$ ($i_t$ is the position that changes from $a_t$ to $b_t$), $y_{i_t} = b_{i_t}$ if $x_t = 1$ and $y_{i_t} = c_{i_t}$ if $x_t = 2$.
Now, it's clear that $(i_1, i_2, \dots, i_{1000})$ is a permutation of $(1, 2, \dots, 1000)$. On the other hand, if $a_t = f(0, 0, \dots, 0, \dots, 0) = (a_1, a_2, \dots, a_{2000})$ and $b_t = f(0, 0, \dots, 1, \dots, 0) = (b_1, b_2, \dots, b_{2000})$, where $t \ge 1000$ is the position of 1, there is a unique $j \ge 1000$ such that $a_j \ne b_j$. Denoting this $j$ by $i_j$, if $x = f(x_1, x_2, \dots, x_{2000}) = (y_1, y_2, \dots, y_{2000})$, one can show in a similar way that $y_{i_t} = a_{i_t}$ if $x_t = 0$ and $y_{i_t} = b_{i_t}$ if $x_t = 1$. Finally, in order to count the number of different functions $f: X \to X$, it suffices to choose $(i_1, i_2, \dots, i_{1000})$, which is a permutation of $(1, 2, \dots, 1000)$, $(i_{1001}, i_{1002}, \dots, i_{2000})$, which is a permutation of $(1001, 1002, \dots, 2000)$, and $(a_{i_t}, b_{i_t}, c_{i_t})$ for $1 \le t \le 1000$ and $(a_{i_t}, b_{i_t})$ for $1001 \le t \le 2000$.
This gives that the total number of functions is $1000!^2 \times 12^{1000}$.