The number of positive integral solutions satisfying the equation ${\mathrm{x}}_1\cdot {\mathrm{x}}_2\cdot {\mathrm{x}}_3\cdot {\mathrm{x}}_4\cdot {\mathrm{x}}_5=240$ is

We have, ${\mathrm{x}}_1\cdot {\mathrm{x}}_2\cdot {\mathrm{x}}_3\cdot {\mathrm{x}}_4\cdot {\mathrm{x}}_5=240$ where ${\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3,{\mathrm{x}}_4,{\mathrm{x}}_5$  are natural numbers.

Now, by using prime factorization of 240, we can write the given equation as ${\mathrm{x}}_1{\mathrm{x}}_2{\mathrm{x}}_3{\mathrm{x}}_4{\mathrm{x}}_5=2^4\times 3\times 5$

Now, we can assign 3 and 5 to any of 5 unknowns in 5 x 5ways.
We can assign $2^4\text{ to }{\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3,{\mathrm{x}}_4,{\mathrm{x}}_5$ in

$\begin{array}{l}\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}=4\\ {\Rightarrow }^{4+5-1}{\mathrm{C}}_{5-1}{=}^8{\mathrm{C}}_4\text{, ways }\end{array}$

$\therefore$ Required number of ways ${=}^8{C}_4\times 25=1750$