Question

# The total number of positive integral solutions $\left(x,y,z\right)$ such that $xyz=24$ is

A

$36$

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B

$45$

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C

$24$

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D

$30$

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Solution

## The correct option is D $30$Explanation of the correct option.Given : $xyz=24$It can be written as $x.y.z={2}^{3}.{3}^{1}$Thus, $x={2}^{{a}_{1}}.{3}^{{b}_{1}}\phantom{\rule{0ex}{0ex}}y={2}^{{a}_{2}}.{3}^{{b}_{2}}\phantom{\rule{0ex}{0ex}}z={2}^{{a}_{3}}.{3}^{{b}_{3}}$We can see that ${a}_{1},{a}_{2},{a}_{3}\in \left\{0,1\right\}$ and ${b}_{1},{b}_{2},{b}_{3}\in \left\{0,1\right\}$Since, ${a}_{1}+{a}_{2}+{a}_{3}=3$Thus, non negative solution$={}^{\left(3+3-1\right)}C_{\left(3-1\right)}$ $={}^{5}C_{2}\phantom{\rule{0ex}{0ex}}=\frac{5!}{2!\left(5-3\right)!}\phantom{\rule{0ex}{0ex}}=10$Since, ${b}_{1}+{b}_{2}+{b}_{3}=1$Thus, non negative solution$={}^{\left(1+3-1\right)}C_{\left(3-1\right)}$ $={}^{3}C_{2}\phantom{\rule{0ex}{0ex}}=\frac{3!}{2!\left(3-2\right)!}\phantom{\rule{0ex}{0ex}}=3$Therefore, total solutions $=10×3$ $=30$Hence $\mathrm{Option}\left(\mathrm{D}\right)$ is the correct option.

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