Question: Question 2 - Lower Bound for Comparison Based Sorts a) Show that log(n!) is O(n log n) b) Show that log(n!) is (n log n)

Question 2 - Lower Bound for Comparison Based Sorts a) Show that log(n!) is O(n log n) b) Show that log(n!) is (n log n) Note: Showing that log(n!) is O(n log n) and log (n!) is (n log n) is equivalent to showing that log(n!) is (n log n) Solution a) (YOUR PART a SOLUTION HERE) b) (YOUR PART b SOLUTION HERE)
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