2. $x_{1}+x_{2}+\cdots+x_{19}+2 x_{20}=3$ has $\qquad$ non-negative integer solutions.
2. 1349 Detailed Explanation: If $x_{20}=1$, then there are $C_{19}^{1}=19$; If $x_{20}=0$, then there are $C_{19}^{1}+A_{19}^{2}+C_{19}^{3}=1330$; Therefore, there are a total of $19+1330=1349$.