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Difficult Math Problem Based on Inversions! Hard! Fun! HELP!

  1. Jun 1, 2010 #1
    1. The problem statement, all variables and given/known data
    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?



    3. 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?!?!?
    :)
     
  2. jcsd
  3. Jun 3, 2010 #2
    If the sixth smallest number is first in the list, then there are 5 inversions due to that number being first in the list. By moving it to the end, we eliminate those 5 inversions. But by moving it to the end, we introduce 14 new inversions, caused by the 7th-20th hands having new inversions. So by moving this number to the right, we have created a net of nine new inversions. Nothing else changes, so our new number of inversions is 47.
     
  4. Jun 3, 2010 #3
    thank you SO much! that makes a lot of sense. i really appreciate it
     
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