In the expansion of ${\left(\sqrt[3]{4}+\frac{1}{\sqrt[4]{6}}\right)}^{20}$

$T_{r+1}{=}^{20}{C}_r{.4}^{\frac{20-r}{3}}\cdot 6^{\frac{-r}{4}}$

${=}^{20}{C}_r{.4}^{\frac{40-2r}{3}}{.2}^{\frac{-r}{4}}{.3}^{\frac{-r}{4}}$

${=}^{20}{C}_r{.2}^{\frac{160-11r}{12}}.3^{\frac{-r}{4}}$

For rational terms, $\frac{r}{4}\text{ and }\frac{160-11r}{12}$ must be integers and $0\le r\le 20.$

So, $\frac{r}{4}=\text{ integer }$

$\Rightarrow r=0,4,8,12,16,20$

$\therefore$Only two terms are rational. So 21 – 2 i.e 19 terms are irrational