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2 votes

yes it is 

this is from Cormen

$it\ is\ \Theta \ since\ when\ an\ algorithm's\ upper\ limit\ and\ lower\ limit\ can\ be\ represented,\ we\ do\ so\ in \ \Theta \\$ 

$ let\ n=4$

$n!=o(n^{n})$

$ 4!<4^{4 }$

$ n!=\omega (2^{n})$

$ 4!>2^{4}$

i hope it’s clear now

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