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# The number of positive integral solutions of the equation $xyz=30$ is ?

$\begin{array}{1 1} 30 \\ 27 \\ 15 \\ 8 \end{array}$

$30=1\times5 \times6$ or $30=1\times2\times15$ or
$30=1\times3\times10$ or $30=2\times 3\times5$ or
$30=1\times\times1\times30$
$\therefore x,y,z$ can take the values from any of the above triplets.
$i.e., 3!\:or\:3!\:or\:3!\:or\:3!\:or\large\frac{3!}{2!}$
The number of solutions=$6+6+6+6+3=27$