61.Take n=3,
Permutation ----> no of inversions
1 2 3 0
1 3 2 1
2 3 1 2
2 1 3 1
3 1 2 2
3 2 1 3
Total inversions =9, no of permutation =6,
So, here n=6
putting it in option n(n-1)/4=6⨉5/4=7.5
So, (B) is most closer option
62. Now elements are comes like 5,4,3,2,1
So, first 5 comes order is 5--------------0 inversion
Now, 4 comes........" " 5,4 -----------1 inversion
Now 3 comes..........." " 4,5,3--------2 inversions
Now 2 "..........................3,4,5,2--------3 inversions
Now 1 " .......................2,3,4,5,1---------4 inversions
total 10 inversion
No more permutation could be done in insertion sort
So, number of permutation could be n(n-1)/2= O(n2)
Ans (A) here