What about (n-1)!?
Also if you could show me a proof that would help me understand better.
I'm stuck on this one.
3 Answers
To show that (n+1)! is in O(n!) you have to show that there is a constant c so that for all big enough n (n > n0) the inequality
(n+1)! < c n!
holds. However since (n+1)! = (n+1) n! this simplifies to
n+1 < c
which clearly does not hold since c is a constant and n can be arbitrarily large.
On the other hand, (n-1)! is in O(n!). The proof is left as an exercise.
(n+1)! = n! * (n+1)
O((n+1)*n!) = O(nn!+n!) = O(2(nn!)) = O(n*n!) > O(n!)
(n-1)! = n! * n-1
O(n-1)! = O(n!/n) < O(n!)
I wasnt formally introduced to algorithmic complexity so take what I write with a grain of salt
That said, we know n^3 is way worse than n, right?
Well, since (n + 1)! = (n - 1)! * n * (n + 1)
Comparing (n + 1)! to (n - 1)! is like comparing n to n^3
Sorry, I dont have proof but expanding the factorial as above should lead to it
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