In order to overlap any previous numbers you have to swap with it atleast once.
Now for every number at pos p you have atleast n-p-1 numbers to be swapped with.

So solution will be 
>! $$\sum_{i=1}^n i = \frac{n*(n-1)}{2}$$ 

Example for 5 4 3 2 1

>! 5 needs to be swapped with 4 numbers  
>! 4 needs to be swapped with 3 numbers  
>! 3 needs to be swapped with 2 numbers  
>! 2 needs to be swapped with 1 number.  
>!   
>! Thus 1+2+3+4 = 10