# How do you prove the parity is the same after the arrangement?

This is not my homework, this is for my recreation and hope you people could solve my trouble.

I have each of ordered sets (x,y) in which x and y equal to a integer from 1 to 5 respectively in which every set have a different x from each set and a different y from each set. I arrange the y number to be from small to big, that is the sequence looks like this:$(x_1,1),(x_2,2),(x_3,3),(x_4,4),(x_5,5)$, and then i want to rearrange those same sets a few time between two set each time in which x will be in a "correct place" to make it become a sequence that looks like:$(1,y_1),(2,y_2),(3,y_3),(4,y_4),(5,y,5)$ for which each x is in correct place now and the rule is we have to make at least one x into "correct place" for each rearrangement. Finally, i get a number $(-1)^p$ for which p is the rearrangement required for the originl sequence to became the "correct place" sequence.

Second of all, i create a chart in ordered that look like this:

1 2 3 4 5

$y_1$$y_2$$y_3$$y_4$$y_5$

for which i connect each number of 1, 2, 3, 4, 5 to a corresponding y to a number that is equal to the y, then i count the number of the intersection of the line which only two line intersect at a point, then for i equal to the intersection number, i compute this number: $(-1)^i$,

prove: $(-1)^p=(-1)^i$

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It may derive from linear algebra, but it's actually a problem in group theory, concerning permutations. – Gerry Myerson Nov 4 '11 at 3:05
@Victor: The title is not the place to tell us about the story of the question (that's what the body of the post is for). The title is to describe the mathematical content (and only the mathematical content) of the question. – Arturo Magidin Nov 4 '11 at 3:27