# Adjacent Pairs In Permutations Both Odd...
Leroy Quet
qqquet at mindspring.com
Mon Nov 4 04:01:42 CET 2002
Just sent this to sci.math:
Consider all of the m! permutations of {1,2,3,4,...,m}.
I THINK (I think that I think) that
the TOTAL number of occurrences of adjacent elements both being odd in
all of the permutations is:
2 * (m-1)! * floor((m-1)/2)
If so, then the number of times (over all permutations) that adjacent
elements are odd would approach an average of once per permutation as m
approaches infinity.
--
Now, the total number of occurrences of any elements being adjacent in
all permutations of {1,2,3,...,m} is:
(m-1) * m!
We can get the same set of permutations by simply subtracting every
element of each permutation from (m+1). If m is even, then the resulting
permutations have elements with parity reversed from the original
permutations.
So the number of occurrences of EVEN elements being adjacent in all of
the permutations of {1,2,3,...,m} is, for EVEN m, the same as for odd
elements:
2 * (m-1)! * floor((m-1)/2)
= (m-1)! (m-2)
So, if I am right, the total number of occurrences of an odd element
being next to an even element (together with the number of occurrences of
an even element being next to an odd element) is, for even m:
(m-1) * m! - 2 * (m-1)! * (m-2) =
(m-1)! (m^2 - 3m + 4)
Half of this is for the number of occurrences of an odd being before an
even.
And (the other...) half is for the number of occurrences of an even being
before
an odd.
So, the number of occurrences of two elements of identical-parity being
adjacent over all permutations of {1,2,3,...m} is of O(m!).
While the number of occurrences of two elements of opposite-parity being
adjacent is of O(m!m).
Therefore, IF true, the number of opposite-parity pairs grows much faster
than the number of same-parity pairs.
But this seems counter-intuitive to me. Is there an easy way to see this
intuitively as fact, if, in fact, it is indeed fact?
Thanks,
Leroy Quet
More information about the SeqFan
mailing list