Answer
The list is already in sorted order. Although the $selection\ sort$ algorithm will go through all of its operations, the largest element will always be at the back of the unsorted section and will be exchanged with itself, resulting in no change.
Work Step by Step
The list is already in sorted order. Although the $selection\ sort$ algorithm will go through all of its operations, the largest element will always be at the back of the unsorted section and will be exchanged with itself, resulting in no change.
