1 votes 1 votes What is the upper bound of n! Please provide a logical deduction to prove that its $n^{n}$. Algorithms algorithms asymptotic-notation + – saxena0612 asked Aug 6, 2017 retagged Jun 22, 2022 by makhdoom ghaya saxena0612 168 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 0 votes 0 votes n! = n*(n-1)*n-2......1 $n^{n}$= n*n*n......n $\frac{n^{n}}{n!}$=$\frac{n*n*n...n}{n*(n-1)*(n-2)*...1}$ This ratio is greater then 1 always for n$\neq 1$ for eg $\frac{3^{3}}{3!} = 4.5$ so $\frac{n^{n}}{n!}\geq 1$ $n^{n}\geq n!$ n! = O($n^{n}$) Tesla! answered Aug 7, 2017 selected Aug 7, 2017 by saxena0612 Tesla! comment Share Follow See 1 comment See all 1 1 comment reply saxena0612 commented Aug 7, 2017 reply Follow Share Got it ! :) 1 votes 1 votes Please log in or register to add a comment.