XXII OBM

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 $1000{!}^2\times 12^{1000}$.