binomial theore tricky question

find the constant term in the expansion of (1 + x )^10(1+ 1/x)^12

$\small{ \begin{array}{|rcll|} \hline && (1 + x )^{10} \left(1+ \dfrac{1}{x} \right)^{12} \\\\ &=& (1 + x )^{10}\left( \dfrac{1+x}{x} \right)^{12} \\\\ &=&\dfrac{(1 + x )^{10}(1 + x )^{12}}{x^{12}} \\\\ &=&\dfrac{(1 + x )^{22}}{x^{12}} \\\\ &=&\dfrac{ \binom{22}{0}x^0 +\binom{22}{1}x^1+ \binom{22}{2}x^2 + \ldots + \binom{22}{11}x^{11}+ \binom{22}{12}x^{12}+ \binom{22}{13}x^{13} + \ldots + \binom{22}{21}x^{21}+ \binom{22}{22}x^{22} }{x^{12}} \\\\ &=& \binom{22}{0}x^{-12} +\binom{22}{1}x^{-11}+ \binom{22}{2}x^{-10} + \ldots + \binom{22}{11}x^{-1}+\binom{22}{12}+ \binom{22}{13}x^{1} + \ldots + \binom{22}{21}x^{9}+ \binom{22}{22}x^{10} \\ \hline \end{array} }$

The constant term is:

$\begin{array}{|rcll|} \hline && \dbinom{22}{12} \\\\ &=& \dbinom{22}{22-12}\\ \\ &=& \dbinom{22}{10} \\\\ &\mathbf{=}& \mathbf{646646} \\ \hline \end{array}$