jmc

To find the coefficient of $a^2{b}^2$ in $(a+b{)}^4{\left(c+\frac{1}{c}\right)}^6$, we need to find the coefficient of $a^2{b}^2$ in $(a+b{)}^4$ and the constant term of ${\left(c+\frac{1}{c}\right)}^6$. Using the Binomial Theorem, we find that these are $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ and $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$. The coefficient of $a^2{b}^2$ in $(a+b{)}^4{\left(c+\frac{1}{c}\right)}^6$ is the product of these, or $\boxed{120}$.