For example, a 1000-bit digital string $a_{1} a_{2} \cdots a_{1000}$, where $a_{i} \in\{0,1\}(i=1,2, \cdots, 1000)$. If two adjacent digits $a_{k} 、 a_{k+1}$ satisfy $a_{k}=1, a_{k+1}=0$, then $a_{k} a_{k+1}$ is called an "inversion" in the digital string. Question: How many 1000-bit digital strings have exactly 300 inversions?
【Analysis】For any thousand-digit number string $x=$ $\overline{a_{1} a_{2} \cdots a_{1000}}$, add the digit 0 at the very beginning and the digit 1 at the very end, to get a 1002-digit number string $y=$, generating 1001 gaps.

Now, add “=” or “$\neq$” in each gap according to the following requirements: if the numbers on both sides of the gap are the same, add “=”, otherwise, add “$\neq$”.

Now, mainly consider the “$\neq$”. Clearly, the numbers on both sides of the 1st, 3rd, 5th, ... “$\neq$” are not in reverse order, while the numbers on both sides of the 2nd, 4th, 6th, ... “$\neq$” are in reverse order.

Since the numbers on both sides of the first and last “$\neq$” are not in reverse order, to get 300 reverse orders, the number string should have exactly 601 “$\neq$”. Conversely, for any $y=\overline{0 a_{1} a_{2} \cdots a_{1000} 1}$, if $a_{1}, a_{2}, \cdots, a_{1000}$ are considered as “undetermined numbers”, and 601 positions are arbitrarily chosen from the 1001 gaps to fill with “$\neq$”, and the remaining positions are filled with “=”, then $a_{1}, a_{2}, \cdots, a_{1000}$ are uniquely determined.

Therefore, “thousand-digit number strings with exactly 300 reverse orders” and “arbitrarily choosing 601 positions from 1001 gaps to fill with “$\neq$”, and the remaining positions to fill with “=”” are in one-to-one correspondence.
Thus, the number of thousand-digit number strings that meet the condition is $\mathrm{C}_{1000}^{600}$.