Number of non negative integer solutions of $x+y+2z=20$ 
The number of non negative integer solutions of $x+y+2z=20$ is

Finding coefficient of $x^{20}$ in
$$\begin{align}
&\left(x^0+x^1+\dots+x^{20}\right)^2\left(x^0+x^1+\dots+x^{10}\right)\\
=&\left(\frac{1-x^{21}}{1-x}\right)^2\left(\frac{1-x^{11}}{1-x}\right)\\
=&\left(1-x^{21}\right)^2(1-x)^{-3}\left(1-x^{11}\right)
\end{align}$$
i.e.finding coefficient of $x^{20}$ in $$(1-x)^{-3}-x^{11}(1-x)^{-3}$$
Or, coefficient of $x^{20}$ in $(1-x)^{-3}-$ coefficient of $x^9$ in $(1-x)^{-3}=\binom{22}{20}-\binom{11}{9}=176$
The asnwer is given as $121$. What's my mistake?
EDIT (after seeing @lulu's comment):
Finding coefficient of $x^{20}$ in $$\begin{align}
&\left(x^0+x^1+...+x^{20}\right)^2\left(x^0+x^2+...+x^{20}\right)\\
=&\left(\frac{1-x^{21}}{1-x}\right)^2\left(\frac{1-x^{22}}{1-x^2}\right)\\
=&\left(1-x^{21})^2(1-x)^{-2}(1-x^{22})(1-x^2\right)^{-1}
\end{align}$$
i.e.finding coefficient of $x^{20}$ in $$(1-x)^{-2}(1-x^2)^{-1}$$
Not able to proceed next.

Found the mistake. Instead of $\left(x^0+x^1+\dots+x^{10}\right)$, it should have been $\left(x^0+x^2+\dots+x^{20}\right)$
