-3
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Knowing that $log(n!) = \theta(nlogn)$

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  • 1
    $\begingroup$ What exactly is the problem? (There is a problem, but can you see it?) $\endgroup$
    – gnasher729
    Commented Jan 4, 2023 at 18:41
  • $\begingroup$ @gnasher729 Yes was a typo, I fixed it $\endgroup$ Commented Jan 4, 2023 at 18:43

1 Answer 1

0
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Hint:

$$\log(mn)\le\color{blue}{\log(mn^2)}\le\log(m^2n^2)=2\log(mn).$$

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