The total number of positive integral solutions (x, y, z) such that xyz = 24 is

$\mathrm{xyz}=24=2^3\times 3^1$

$\mathrm{Let} \mathrm{x}=2^{{\mathrm{\alpha }}_1} 3^{{\mathrm{\beta }}_1} , \mathrm{y}=2^{{\mathrm{\alpha }}_2} 3^{{\mathrm{\beta }}_2} , \mathrm{z}=2^{{\mathrm{\alpha }}_3} 3^{{\mathrm{\beta }}_3}$

$\Rightarrow \mathrm{xyz}=2^3\times 3^1=2^{{\mathrm{\alpha }}_1+{\mathrm{\alpha }}_2+{\mathrm{\alpha }}_3} 3^{{\mathrm{\beta }}_1+{\mathrm{\beta }}_2+{\mathrm{\beta }}_3}$

$\Rightarrow {\mathrm{\alpha }}_1+{\mathrm{\alpha }}_2+{\mathrm{\alpha }}_3=3, {\mathrm{\beta }}_1+{\mathrm{\beta }}_2+{\mathrm{\beta }}_3=1$

$\begin{array}{rcl} \mathrm{Number} \mathrm{of} (\mathrm{x}, \mathrm{y}, \mathrm{z}) & =& \left({}^{3+3-1}\mathrm{C}_{3-1}\right) \left({}^{1+3-1}\mathrm{C}_{3-1}\right)\\ & =& {}^5\mathrm{C}_2 \left({}^3\mathrm{C}_2\right)=\\ & =& 10\left(3\right)=30\end{array}$