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QUESTION Explain why the pictorial way of counting inversions
gives the correct answer.


ANSWER \setlength{\unitlength}{.5in}
\begin{picture}(8,4)
\put(2,0.8){r}\put(1,3.2){r}\put(3,0.8){s}\put(3,3.2){s}
\put(5,3.2){r}\put(7,3.2){s}\put(6,0.8){s}\put(7,0.8){r}
\put(2,1){\line(-1,2){1}} \put(3,1){\line(0,1){2}}
\put(6,1){\line(1,2){1}}\put(7,1){\line(-1,1){2}}
\end{picture}

Consider a pair $(r,s)$ with $r<s$. The above pictures illustrate
typical cases where $r$ precedes $s$ in the permutation
(non-inversion) and where $s$ precedes $r$ (inversion). the
corresponding lines in the picture do not cross in the former case
but do in the latter.


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