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katek8k8
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Homework Statement
Given integers 4, 8, 7, 5. An inversion occurs when one integer is larger than an integer that follows it; thus, in the above arrangement there are two inversions due to 8 and one due to 7 or three inversions in all. There are twenty distinct possible integers arranged in some order with the sixth smallest in the first position. The number of inversions in this arrangement in 38. If the first integer is now moved to the last position, how many inversions are there in the new arrangement?
The Attempt at a Solution
Well if there are 38 inversions, and the sixth smallest number is first, that means there has to be 15 inversions following it in the pattern that would be eliminated by 5 if it were to move to the end of the list of numbers. That leads us to 33 inversions, but the problem still isn't complete.
Someone help PLEASE?
:)