11. The sequence $\left\{a_{\mathrm{n}}\right\}$ has 1001 terms, $a_{1}=0, a_{1001}=2020$, and $a_{\mathrm{k}+1}-a_{\mathrm{k}}=1$ or $3, \mathrm{k}=1,2, \ldots, 1000$. Then the number of different sequences that satisfy this condition is
(use a combination number as the answer);
Solution: Given $a_{\mathrm{k}+1}-a_{\mathrm{k}}=1$ or $3, a_{1001}=\left(a_{1001}-a_{1000}\right)+\left(a_{1000}-a_{999}\right)+\cdots+\left(a_{2}-a_{1}\right)=2020$, let the 1000 differences $a_{1001}-a_{1000,} a_{1000}-a_{999}, \ldots, a_{2}-a_{1}$ contain $\mathrm{x}$ ones and $\mathrm{y}$ threes, then we have $\left\{\begin{array}{l}x+3 y=2020 \\ x+y=1000\end{array}\right.$, solving this gives $\left\{\begin{array}{l}x=490 \\ y=510\end{array}\right.$, meaning that among the 1000 differences $a_{\mathrm{k}+1}-a_{\mathrm{k}}(1 \leqslant \mathrm{k} \leqslant 1000)$, there are 490 ones and 510 threes. Since each permutation of these 490 ones and 510 threes uniquely corresponds to a sequence that meets the conditions, the number of such sequences is $\frac{1000!}{490!510!}=C_{1000}^{490}$.