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Combinatorics of Permutations, 2nd Edition
book

Combinatorics of Permutations, 2nd Edition

by Miklos Bona
April 2016
Intermediate to advanced content levelIntermediate to advanced
478 pages
14h 44m
English
Chapman and Hall/CRC
Content preview from Combinatorics of Permutations, 2nd Edition
120 Combinatorics of Permutations, Second Edition
PROOF By Theorem 3.53, we need to compute
G
C
(x)=exp
x +
x
3
3
+
x
5
5
+ ···
=exp
n0
x
2n+1
(2n +1)!
.
Now note that taking derivatives, we have
n0
x
2n+1
2n +1
=
n0
x
2n
=
1
1 x
2
=
1
2(1 x)
+
1
2(1 + x)
.
Therefore,
n0
x
2n+1
2n +1
=
1
2
ln
(1 x)
1
+ln(1+x)),
and so
G
C
(x)=exp
1
2
ln
(1 x))
1
+ln(1+x)
=
1+x
1 x
,
which was to be proved.
COROLLARY 3.59
For all positive integers n, the number of permutations of length 2n that
have odd cycles only is ODD(2n)=(1· 3 · 5 ···(2n 1))
2
=(2n 1)!!
2
.
Similarly, the number of permutations of length 2n +1 that have odd cycles
only is ODD(2n +1)=(1· 3 · 5 ···(2n 1))
2
(2n +1)=(2n
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Publisher Resources

ISBN: 9781439850527