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Let $k$ and $n$ be integers greater than $1$. Then $(kn)!$ is not necessarily divisible by

1. $(n!)^k$
2. $(k!)^n$
3. $n! \cdot k! \cdot$
4. $2^{kn}$

recategorized | 21 views

$D$

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.
by Boss (19.1k points)

+1 vote