Let $\mathbb{S}=\{1,\ 2,\ \cdots,\ 1000\}$ and $\mathbb{A}$ be a subset of $\mathbb{S}$. If the number of elements in $\mathbb{A}$ is $201$ and their sum is a multiple of $5$, then $\mathbb{A}$ is called $\textit{good}$. How many good $\mathbb{A}$ are there?
Dividing all subsets of $\mathbb{S}$ containing $201$ elements into $5$ groups: $\mathbb{S}_0$, $\mathbb{S}_1$, $\mathbb{S}_2$, $\mathbb{S}_3$, and $\mathbb{S}_4$ according to the sum of their elements modulo $5$. For example, all subsets having $201$ elements whose sums are multiples of $5$ are in $\mathbb{S}_0$, whose sums are multiples of $5$ plus $1$ are in $\mathbb{S}_1$, and so on.
Now, we claim that $\mathbb{S}_0$ and any of $\mathbb{S}_k$, $k=1,\ 2,\ 3,\ 4$, have a bijective relation. This is because for every $a_i\in\mathbb{S}_0$, there is a unique element $b_i\in\mathbb{S}_k$ satisfying $b_i = a_i + k \pmod{1000}$.
It follows that counts of all these five groups are equal. Meanwhile, the total counts of these five groups is $\binom{1000}{201}$ because this value is equivalent to choosing $201$ elements from a $1000$ choices in $\mathbb{S}$. Therefore, the number of good $\mathbb{A}$, i.e., $\mathbb{S}_0$, is $$\boxed{\frac{1}{5}\binom{1000}{201}}$$