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Prove that $n!=\omega(2^n)$ and $n!=o(n^n)$.
in Algorithms 43 views

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n! = omega(2^n)

n! = n * (n-1) * (n-2) * .......2 * 1     >      2 * 2 * 2* 2*  ...... * 2 (n times)          ; n > 2, (n-1) > 2 , (n-3) >2

so, n! =  omega(2^n)

n! = n * (n-1) * (n-2) * .......2 * 1    <     n * n * n *n         *n   (n times)           ; n = n , n-1 <n ; n-2 <n ; n-3 < n

so , n! = big-O( n^n)

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