1 votes 1 votes Let $k$ and $n$ be integers greater than $1$. Then $(kn)!$ is not necessarily divisible by $(n!)^k$ $(k!)^n$ $n! \cdot k! \cdot$ $2^{kn}$ Quantitative Aptitude isi2015-mma quantitative-aptitude number-system remainder-theorem + – Arjun asked Sep 23, 2019 • recategorized Nov 16, 2019 by Lakshman Bhaiya Arjun 656 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes $D$ Let $k=n=2$ Now, $(kn)! = (2*2)! = 4! = 24$ which is not divisible by $2^{kn} = 2^4 = 16 $ Hence, option $D$ is the right answer. `JEET answered Sep 24, 2019 `JEET comment Share Follow See all 0 reply Please log in or register to add a comment.