# Sort Complexities
## Aug. 17, 2016
Question
Sort the following complexities:
e^nen
n!n!
n^{100}n100
n^nnn
Hint
Think of what happens whennnis extremely large.
If in doubt, try to break the term into multiple multiplication steps.
Solution
It is pretty clear thatn^{100} < n^nn100 100n>100. It is also clear thate^n < n^nen en>e.
What is fuzzy for many is where shouldn!n!fit in. Let us break the complexities down into smaller terms:
en=e×e×e×…×e×en!=n×n−1×n−2×…×2×1nn=n×n×n×…×n×nen=e×e×e×…×e×en!=n×n−1×n−2×…×2×1nn=n×n×n×…×n×n
As you can see above, asnngrows,n!n!easily overtakese^nen. Therefore, the answer is\enspace n^{100} < e^n < n! < n^nn100
Below is a useful Big-O Complexity Graph for you to visualize how the number of operations grow asnnincreases.
