JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Permutations and Combinations

Writing $24=2^3\cdot 3$, an ordered triple $(x,y,z)$ with $xyz=24$ is fixed by sharing the three factors of $2$ and the single factor of $3$ among the three places. The exponent $3$ can be split among three places in $\left(\genfrac{}{}{0ex}{}{3+2}{2}\right)=10$ ways and the exponent $1$ in $\left(\genfrac{}{}{0ex}{}{1+2}{2}\right)=3$ ways, and the two splittings are independent. The number of positive integral solutions is $10\times 3=30$.