Question

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

• A. $36$
• B. $45$
• C. $24$
• D. $30$

Answer 1

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,

$$\begin{gathered} x=2^{a_1}.3^{b_1} \\ y=2^{a_2}.3^{b_2} \\ z=2^{a_3}.3^{b_3} \end{gathered}$$

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$={}^{(3+3-1)}C_{(3-1)}$

$$\begin{gathered} ={}^{5}C_2 \\ =\frac{5!}{2!(5-3)!} \\ =10 \end{gathered}$$

Since, $b_1+b_2+b_3=1$
Thus, non negative solution$={}^{(1+3-1)}C_{(3-1)}$

$$\begin{gathered} ={}^{3}C_2 \\ =\frac{3!}{2!(3-2)!} \\ =3 \end{gathered}$$

Therefore, total solutions $=10\times 3$

$=30$

Hence $\mathrm{Option}\left(\mathrm{D}\right)$ is the correct option.