complexity of judging n numbers is sorted or not

my analysis:
n! permutations.
i times comparation for a permutation.
i=1: 1*sikma(k=n to 2)* C(k-1,1)*(n-i-1)! comparations
...
i=t: t*sikma(k=n to i+1)*C(k-1,t)*(n-t-1)!
...
i=n-1: (n-1)*sikma(k=n to n)*C(k-1,n-1)*(n-(n-1)-1)!=n-1

so the answer is:

sikma(t=1 to n-1)[ (t*sikma(k=n to i+1)*C(k-1,t)*(n-t-1)! ) ] / n!

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